Matematika Ii Dio

Viša tehnička škola Doboj

Vježbe Prof.Vesna Mišić

Doboj 2003 1

Matematika

Funkcija - definicija 2

Definicija : Funkcija je preslikavanje skupa A u skup B pomoću koga svakom elementu skupa A pridružujemo tačno jedan element skupa B. preslikava f : A  → B

Domen ? f( x) 1. y = g( x) 2. y =

f

A x

D : g( x) ≠ 0

f( x)

f

D : f( x) ≥ 0

B y

−1

Domen (područje definisanosti), Kodomen (skup vrijednosti ).

3. y = log f ( x ) D : f ( x ) > 0 4. y = arcsin x D : − 1 ≤ x ≤ 1

1 y=sinx 0 -1

π

2π

y=arcsinx

Inverzne funkcije. f : A→B f : x→B x 0 30 si nx

° 0

° 1 2

x

0

arcsinx

0

3

f

−1

:B→ A

f

−1

:y→x

45° 60° 90 ° 2 3 1 2

1 80° 0

2 70° -1

360° 0

2 1 2 π 6

2 2 π 4

3 2 π 3

1

0

-1

0

π 2

π

3π 2

2π

Funkcija f : A → B je : 1. Injektivna (1-1) ako f ( a ) = f ( b ) ⇒ a = b . 2. Sirjektivna (na) cio skup b ako je f ( A ) = B . 3. Bijektivna (ako je injektivna i sirjektivna). 4. Parna ako je f ( − x ) = f ( x ) za svako x iz A . 5. Neparna ako je f ( − x ) = − f ( x ) za svako x iz A ( ∀x ∈ A) , (simetrična u odnosu na koordinatni početak) . 6. Periodična s periodom ω, ako vrijedi f ( x +ω ) = f ( x ) ( ∀x ∈ A) . 7. Rastuća x1 < x 2 ⇒ f ( x1 ) < f ( x2 ) ( ∀x1 , x 2 ∈ A) . 8. Opadajuća x1 < x 2 ⇒ f ( x1 ) > f ( x2 ) ( ∀x1 , x 2 ∈ A) . 9. Ograničena m ≤ f ( x ) ≤ M ( ∀x ∈ A) , ( m < ∞, M < ∞ ) . 1. a b

1 4 2 5

c

3

Injektivna, ali nije sirjektivna 2. A

1

B 0

2

Sirjektivna ali nije, injektivna, f

3.

2 4

1 2

6 8 10

3 4 5

Bijektivna (injektivna i sirjektivna) 4. Parna

y=cosx 0 4

y=x2

5.

y=x

y=x3

6. y=sinx

7.

f ( x2 )

) f( ′ ) x f(

x2′ 1

x1 x2 a

Domen. 5

f ( x1 )

x1′ x2 ′

b

1. y = sin ln( 2 x − 3)

x x −1 + x2 − 4 2 x+3 ( x − 4) + 4 − 3 x − x 2 3. y = ln x+2

2. y = 2 sin + ln

4. y = sin π ( x − 1) + 4 − x 2 1. D : 2 x − 3 > 0 ⇒ x >

3 2

x 2

2. 2 sin = 0 ∀x ∈ R x −1 >0 x + 3 = 0 ∧ x −1 > 0 x+3 x = −3 ∧ x > 1 -∞

-3

1

∞

x

+

x-1 x+3 x −1 x+3

- 0 + 0 + + +

I1 D : x ∈ ( − ∞ ,− 3) ∪ ( 1,+∞ ) x2 − 4 ≥ 0 ( x − 2)( x + 2) ≥ 0 x1 = 2 ∧ x 2 = −2

-∞

x+2 x2-4

I 2 D : x ∈ ( - ∞ ,-2] ∪ [ 2,+∞ )

6

2

x x-2

( D : I1 ∩ I 2 )

-2

+

- 0 + 0 + + +

∞

-3

-2

1

2

D : x ∈ ( − ∞,−3) ∪ [ 2,+∞ ) x−4 x−4 > 0 ∧ ln ≥0 x+2 x+2

3. D :

x−4=0 x=4 x+2=0 x = −2

x−4− x−2 ≥0 x+2 −6 ≥0 x+2 x+2<0 I 2 : x < −2 D : x ∈ ( − ∞ , −2 )

x−4 ≥1 x+2 x−4 −1 ≥ 0 x+2

-∞

-2

4

∞

x

-

x-4

+

x+2 x−4 x+2

- 0 + 0 + + +

I1 D : x ∈ ( - ∞ ,-2) ∪ ( 4,+∞ ) 4 − 3x − x 2 ≥ 0 3 ± 9 + 16 x1, 2 = −2 x1 = 1

− x 2 − 3x + 4 ≥ 0

x 2 = −4

I 3 : D : x ∈ ( − 4,1)

D : x ∈ [ − 4,− 2) -4

7

-2

1

4

4. y = sin π ( x − 1) + 4 − x 2 sin π ( x − 1) ≥ 0 0 ≤ π ( x − 1) ≤ π 0 ≤ x −1 ≤ 1

x ≥ 1 ∧ x ≤ 2 I1′ − 4π ≤ π ( x − 1) ≤ −3π − 4 ≤ x − 1 ≤ −3 − 3 ≤ x ≤ −2

-3

− 2π ≤ π ( x − 1) ≤ −π / : π − 2 ≤ x − 1 ≤ −1 x ≥ −1 ∧ x ≤ 0 I ″ 1

/ :π

-2

D : x ∈ { 2} ∪ [ − 1,0] ∪ [1,2] 5. y = x − 3 − x x−3 = 0

-1

0

x ∈ ( − ∞ ,0 )

I1

1

y = −( x − 3) − x y = −2 x + 3

y = −( x − 3) + x y=3

x =0 I1

I2

( 3,+∞ )

y = x −3− x y = −3

x y

0

2

3

-1

I3

0 I3

2

x ∈ [ 0,3)

I2

y = x−3 − x

x =3

4 − x2 ≥ 0 ( 2 + x )( 2 − x ) ≥ 0 I 2 x ∈ [ − 2,2]

3 y=3 3 y=-2x+3 3 y=-3 -3

1) Da li je funkcija injektivna? y = 3x + 5

5

−

8

5 3

0

Sirjektivna

2) Naći nule funkcije

(

y = ln x 2 − 3

)

D : x2 − 3 > 0 y=0 ln  1 = 0

(

) (

D : x ∈ − ∞ ,− 3 ∪

3 ,+∞

)

0 = 1 f = 0 ⇒ x2 − 3 = 1 x2 = 4 x1, 2 = ±2

-2

y>0

x ∈ ( − ∞ ,− 2) ∪ ( 2,+∞ )

y<0

x ∈ − 2,− 3 ∪

(

) (

3 ,2

− 3

3

2

)

8. Koja je funkcija parna, a koja je neparna? y = x 3 + 3x + 1 funkcija u tački − x y ( − x ) = ( − x ) 3 + 3 ⋅ ( − x ) + 1 = − x 3 − 3 x + 1 nije ni parna ni neparna. x + − x 2 ( −x)  + −( − x ) x + − x y( − x ) = = = y ( x ) Funkcija je parna. 2 2 x 9. y = ln Inverzna funkcija. 4 y y x = ln  x = 4x = y 4 4 funkcija siječe osu y u broju 4. 1 10. y = 1 − x − 1 y( x ) =

I1 D : x − 1 = 0 9

x =1

x ∈ ( − ∞,1)

1 1 + ( x − 1) 1 y= D : x ∈ R /{ 0} x y=

I2

y

x ∈ [1,+∞ )

x

1 1 − ( x − 1) 1 y= 2− x D: x ≠ 2 y=

x=0

1 D: x ≠3 11. y = x−3 x > 0 funkcija je pozitivna x < 0 funkcija je negativna

y=−

x=1

x=2

1 3

x=3

12. y =

1 2−x

D: x ≠ 2

x > 0 funkcija je negativna x < 0 funkcija je pozitivna 1 x = 0 siječe y – osu u y = 2

y=

1 2

x=2

Granične vrijednosti - ( limesi ) 1 an = - niz n

0

1 4

1 1 3 2

1

(a-ε, a+ε) - epsilon okoline tačke a 10

a-ε

(

a

)

a+ε

ε okoline tačke a

1

=0 lim n→∞ n

a 0 n k + a1n k −1 + ⋅ ⋅ ⋅ + a n ⋅ n 0

lim k k −1 + ⋅⋅⋅ + b n →∞ b n + b n

1 =∞ lim n →0 n

0

1

n

⋅n

0

=

a0 b0

( k ∈ N )( n ≠ 0 ∧ k ≠ 0)

7 6n − 7 6n − 7 / : n n =6=2 = lim = lim 13. lim 3 n →∞ 3n + 2 n→∞ 3n + 2 / : n n →∞ 3 + 2 n n  1  1  1  1  ⋅  n − 1     − 1 1− n  1  1  3 1 1  = 3 =  3 3 14. lim  + 2 + ⋅ ⋅ ⋅ + n  = lim  ⋅ 1 lim lim  2  3 3 2  n →∞ 3 3  n →∞ n →∞  − 1  n →∞  −     3 3      6−

  = 1  2  

15. 2 2n +4 (1 + 2 n ) 1 + 2 + 3 + ⋅ ⋅ ⋅ + 2n 2 n ( 1 + 2 n ) 2 + 4 n n n 2 = = = : = =4 lim lim lim lim n 1 1+ 2 + ⋅⋅⋅ + n n( 1 + n ) 1 + n n lim n →∞ n →∞ n → ∞ n → ∞ n → ∞ (1 + n ) +1 2 n x 2 − 5 12 − 5 4 = =− 16. lim 1+ 2 3 x →1 x + 2

17.

x 2 + 6 x − 16 ( x − 2)( x + 8) = x+8 2+8 = lim = =2 lim lim 2 2+3 x + x−6 x →2 x →2 ( x − 2 )( x + 3) x→2 x + 3

18.

(

)

2  3 ⋅ (1 + x ) − 2 ⋅ 1 + x + x 2 3 + 3x − 2 − 2 x − 2 x 2  3 − = = =   lim lim lim 3 2 1 − x 2  x→1 (1 − x ) ⋅ (1 + x ) ⋅ 1 + x + x 2 x →1  1 − x x →1 (1 − x ) ⋅ (1 + x ) ⋅ 1 + x + x

(

)

(

)

1 + x − 2x ( − 2 x − 1)( x − 1) − ( − 2 x − 1) = lim = lim = 2 2 2 x →1 (1 − x ) ⋅ (1 + x ) ⋅ 1 + x + x x →1 − (1 − x ) ⋅ (1 + x ) ⋅ 1 + x + x x →1 (1 + x ) ⋅ 1 + x + x 2x + 1 2 +1 1 = lim = = 2 2⋅3 2 x →1 (1 + x ) ⋅ 1 + x + x 2

= lim

(

(

)

(

)

)

x2 x = −1 - vertikalna asimptota x +1 y = 0 ⇒ x = 0 nula funkcije je u nuli y 2 x2 y x2 x2 1 : 2 = lim =1 kosa asimptota k = lim = lim x + 1 = lim 2 x x →∞ x x →∞ x →∞ x + x x x →∞ 1 + 1 x 2 2 2  x  x − x −x x −x  = lim n = lim ( y − k ⋅ x ) = lim  − x  = lim  = −1  x + 1  x →∞ x + 1 x →∞ x →∞  x + 1 x →∞   0 -1

19. y =

x y= x+1

11

(

)

x2 − 4 20. y = 2 x −1

y( − x ) =

( − x) 2 − 4 = ( − x) 2 − 1

x2 − 4 = y funkcija je parna x2 −1

x 2 − 4 = 0 ⇒ x1, 2 = ±2 nule funkcije x 2 − 1 = 0 ⇒ x = ±1 vertikalne asimptote x2 − 4 1− 4 − 3 = = =∞ lim 2 1−1 0 x →1 x − 1 x 2 − 4 ( − 1) 2 − 4 − 3 = = =∞ lim 2 ( − 1) 2 − 1 0 x → −1 x − 1

vertikalne asimptote

4 1− 2 x2 − 4 x2 − 4 / : x2 x =1 = lim 2 = lim lim 2 2 1 x →+∞ x − 1 x →+∞ x − 1 / : x x →+∞ 1 − x2

y x =0⇒ y =

0−4 =4 0 −1

(0,4)

y=1 -1 0 1 2 −x x -2 x y( − x ) = = − ( − x ) 2 − 9 x 2 − 9 funkcija je neparna x2 − 9 D : x ≠ ±3 x = 0 nula funkcije

21. y =

12

x

x

0

y = lim 2 = = 0 horizontalna asimptota je x – osa lim 1 x →∞ x →∞ x − 9 x

3

3

= lim = =∞ lim 2 0 x →3 x − 9 x →3 9 − 9

vertikalne asimptote

x −3 −3 = lim = =∞ lim 2 2 0 x → −3 x − 9 x → −3 ( − 3) − 9 y

y=

-4

-3

-2

-1

0

1

2

3

4

x x2 − 9

x

x=3

x=-3

OMM X ~ OAB 5 2 proporcionalnost = ∞; =0 0 ∞ stranica 0 ∞ ∞ 0 ; ; ; ; neodređeni oblici MM X AB 0 ∞ 0 ∞ = B OM X OA 1 1− 3 x 1M− 3 x 1 1 = lim − = −3 =− 22. lim x − 1 = lim 3 3 3 2 2 3 1 + 1 +1 x →1 x →1 − 1 − 3 x x + 3 x + 1 sinxα→1 ABx + 3 x + 1 = ⇒ AB = tgα α cos α 1 4 + x + 2 ⋅  ( 4 + x ) 2 + 3 ( 4 + x ) ⋅ 4 + 3 16  4+ x −2 0 4 +Mxx − 2A(1,0)  = = lim 3 ⋅ lim 3 3 3 2 4+ x − 4 4+ x − 4 x →0 x →0 4 + x + 2 ⋅  ( 4 + x ) + 3 ( 4 + x ) ⋅ 4 + 3 16    23. ( 4 + x − 4) ⋅  ( 4 + x ) 2 + 3 ( 4 + x ) ⋅ 4 + 3 16  3 16 + 3 16 + 3 16 3 ⋅ 23 2 33 2   = lim = POAM= < Pisj.OAM < POAB = 2+2 4 2 ( 4 + x − 4) ⋅ 4 + x + 2 x →0 1 ⋅ sin x x ⋅ 1 1 ⋅ tgx sin x < < /: B 2 2 2 2 M x 1 sin x 1< < = 1 24. lim x sin x cos x x →0 sin x cos x < <1 Mx A Dokaz : 0 x x → 0 ⇒ cos x = 1 sin x 1< <1 x 13 sin x =1 lim x x →0

(

)(

(

( (

)

)

) )

sin ( α + β ) = sin α cos β + cos α sin β sin ( α − β ) = sin α cos β − cos α sin β sin ( α + β ) − sin ( α − β ) = 2 sin α cos β α+β = x α+β = x (+) (-) α −β = y α −β = y 2α = x + y 2β = x − y x+ y x− y α= β= 2 2 sin x − sin y = 2 cos

14

x+ y x− y ⋅ sin 2 2

25.

(

)(

)

sin 4 x − sin 4 a sin 2 x − sin 2 a ⋅ sin 2 x + sin 2 a = lim = lim x−a x−a x→a x→a = lim

( sin x − sin a )( sin x + sin a ) ⋅ (sin 2 x + sin 2 a ) =

x−a x+a x−a 2 ⋅ cos ⋅ sin ⋅ ( sin x + sin a ) ⋅ sin 2 x + sin 2 a 2 2 = lim = x−a x →a x+a x−a cos ⋅ sin ⋅ ( sin x + sin a ) ⋅ sin 2 x + sin 2 a 2 2 = lim = cos a ⋅ 2 sin a ⋅ 2 sin 2 a = 4 ⋅ sin 3 a ⋅ cos a x − a x →a 2 0 ∞ 1∞;0 ∞; ∞0 ; ; Neodređeno 0 ∞ x →a

(

)

(

x

 1 1 +  =  lim x x →∞  ln(1 + x ) =1 lim x x →0

1

(1 + x ) x lim x→0

)

=   ≈ 2,71828

a x −1 = ln a lim x x →0

x − 1 =1 lim x x →0

lim x →0

(1 + x ) a − 1 = a x

26. x

x +1 −2 ⋅ ⋅x x +1

x

− 2  −2  x −1  x +1− 2     = lim   = lim 1 +  lim x + 1 x →∞  x + 1  x →∞  x + 1  x →∞  =

− 2x / : x

= lim lim x →∞ x + 1 / : x x →∞

x + 1 27. lim   x →∞  x 

28.

x2

(1 + sin x ) lim x →0

−2 1 = −2 = − 2 = 2 1  1+ x

 1 = lim 1 +  x x →∞  1 x

=

x⋅ x

= lim x = ∞ = ∞ x →∞

1

1 x x  = lim 1 + sin x ⋅  = lim (1 + x ) x =  sin x  x →0  x →0

29. 1

( cos x ) x lim x →0 =

lim x →0

2

1

= lim (1 + cos x − 1) cos x −1⋅

cos x −1 x2

=

x →0

cos x − 1 − (1 − cos x ) = lim = − lim 2 x x2 x →0 x →0

2 sin 2 x2

x x 1 sin 2 2 =− 2 = − 1 = − 2 = 1 lim 2  x →0 2 ⋅ x ⋅ x 2 2

II. način sin 2

15

x 1 − cos x = 2 2

2 sin 2

x = 1 − cos x 2

cos x = 1 − 2 sin 2

x 2

1

1

2   2 xx 2 x  − 2 sin 2 x 1 − 2 sin = 1 + − 2 sin     2 lim lim 2 2 x →0  x →0 

⋅

− 2 sin 2 x2

x 2

−

=

1 2

1

=



1

 1  x − 1   x − 1 = x ⋅ =1 30. lim   lim 1 x ←∞ x → ∞   x

Odrediti asimptote funkcija : 1 ⇒ x D: x ≠ 0

31. y = x −

lim x →0

2

y=

x −1 = lim x x →0

y = kx + n

x2 −1 x 1 x 2 = 1 = ∞ vertikalna asimptota je x=0 tj. y – osa. 1 ∞ x

1−

1 x2 −1 1− 2 2 x −1 x =1 k = lim x = lim 2 = lim x 1 x x →∞ x →∞ x →∞ 1 −  x2 −1  x2 −1− x2 n = lim  − x  = lim = x =0 x x 1 x →∞   x →∞ y = x kosa asimptota

nule funkcije x 2 − 1 = 0 x2 = 1 x1, 2 = ±1 y y=x

y = x−

1 x

x

x=0 D: x+2≠ 0 x−2 32. y = vertikalna asimptota x = −2 x+2 x ≠ −2 x − 2 = 0 ⇒ x = 2 nula funkcije

16

2 x−2/:x x =1 = lim lim x →∞ x + 2 / : x x →∞ 1 + 2 x 1−

y = 1 horizontalna asimptota

Znak -∞

-2

x-2

-

x+2 f(x)

2

+

0

+

+∞

0

+ +

-

+

y=1 -2

-1

0

1

2

y=

(0,-1)

33. y =

x

x−2 x+2

D : x + 1 ≠ 0 x ≠ −1

( x + 1) 2 x

−1

x

1

= lim 2 = lim = − = ∞ lim 2 0 x →−1 ( x + 1) x → −1 x + 2 x + 1 x → −1 1 − 2 + 1

vertikalna asimptota

1 x 0 x = = = 0 horizontalna asimptota je y = 0 x- osa lim lim 2 1 x →∞ x + 2 x + 1 x →∞ 1 + 2 + 1 x x2 y>0 x>0 znak y < 0 x < 0 y′ =

x 2 + 2 x + 1 − x ⋅ ( 2 x + 2)

y ( 1) =

17

( x + 1)

1

(1 + 1) 2

=

4

=

x 2 + 2x + 1 − 2x 2 − 2x

( x + 1)

4

=

− x2 +1

( x + 1)

4

=

1  1 1,  funkcija ima maksimum u tački T 4  4

(1 + x )(1 − x ) = 1 − x ( x + 1) 4 ( x + 1) 3

y

y=

x

( x +1) 2

1 4 -1

0

1

y=0

x

x=-1

34. y = x − x 2 + x − 2

D : x 2 + x − 2 ≥ 0 ⇒ x ∈ ( − ∞,−2] ∪ [1,+∞ )

2 lim x − x + x − 2 = lim

(x −

)(

x2 + x − 2 ⋅ x + x2 + x − 2

x + x2 + x − 2 2 −1+ −x+2 1 x = lim = lim =− 2 2 1 2 x →∞ x + x + x − 2 x →∞ 1+ 1+ − 2 x x x →∞

x →∞

) = lim x x →∞

2

− x2 − x + 2

x + x2 + x − 2

y = 0 x − x2 + x − 2 = 0 x = x2 + x − 2 /2

x = 2 nula funkcije

x2 = x2 + x − 2

y

y = x − x2 + x − 2 nema funkcije -2

-1

(-2,-2)

18

0

x 1

2

y=−

1 2

=

x2 +1 x2 − 4 D : x2 − 4 ≠ 0

35. y =

x1, 2 ≠ ±2 x ∈ ( − ∞,−2 ) ∪ ( − 2,2) ∪ ( 2,+∞ )

x 2 + 1 = 0 nema nula ( − 2) 2 + 1 = 5 = ∞ x2 +1 = lim lim 2 2 0 x → −2 x − 4 x → −2 ( − 2 ) − 4

vertikalne asimptote x = −2 ∧ x = 2 x2 +1 22 + 1 5 = lim 2 = =∞ lim 2 0 x →2 x − 4 x →2 2 − 4 1 1+ 2 2 x +1 x =1 = lim horizontalna asimptota y = 1 lim 2 x →∞ x − 4 x →∞ 1 − 4 x2 Znak -∞

-2

x2 +1

+

x−2

-

x+2

-

y

(

)

(

0

) = 2x

2x ⋅ x 2 − 4 − 2x ⋅ x 2 + 1

Znak y ′

(x

2

−4

)

2

3

(x

-10x

x 4 − 8 x 2 + 16

19

x2 +1 0 +1 1 = =− 2 4 x −4 0−4

2

+ +

0

+

+

-

+

− 8x − 2x 3 − 2 x

-

x = 0 y( 0) =

+∞

+

+

Prvi izvod funkcije y′ =

2

−4

)

2

-2

− 10 x x 4 − 8 x 2 + 16

=

0

2

+

+

+

-

-

+

+

+

+

+

+

-

-

x−4 2 =y x+1

2

y

y=1 x -2

-1

0

1

2

T

ln x + 2 ln x D : ln x ≠ 0 ⇒ x > 1

x= -2

x=2

36. y =

1 ln x + 2 y = lim = lim x = 1 Horizontalna asimptota y =1 lim ln x x →∞ x →∞ x →∞ 1 x ln x + 2 0 + 2 2 y = lim = = = ∞ Vertikalna asimptota je x=1 lim ln x 0 0 x →1 x →1 1 ln x + 2 y = lim = lim x = 1 lim ln x x →0 x →0 x →0 1 x y = 0 ln x + 2 = 0 ln x = −2 x = − 2 1 x= 2 

Znak funkcije

20

-∞

1

ln x + 2

-

ln x

-

-

y

+

-

0

+∞

+

+

0

+ +

( ,3)

3

y=

ln x + 2 ln x

( 0,1) y=1

1   2 ,0   

1



x

x=1

Naći nule funkcija ? x3 − 8 x3 + 8 D : x3 + 8 ≠ 0

37. y =

( x + 2) ( x 2 − 2 x + 4 ) ≠ 0

x+2≠0 ∧ x ≠ −2

x 2 − 2x + 4 ≠ 0

( x − 2) ( x 2 + 2 x + 4) = 0 x=2

Nula funkcije je x = 2 x 4 − 17 x 2 + 16 x2 + 2 D : x2 + 2 ≠ 0

38. y =

(∀

x ∈ R)

x2 = t t 2 − 17t + 16 = 0

t1 = 1 t 2 = 16

x1, 2 = ±1

∧

39. y = log( x − 1)

x −1 > 0 x >1 D: log( x − 1) = 0 x −1 = 1

21

⇒

x2 = 1

x3, 4 = ±4

∧ x 2 = 16 y

y = log( x − 1)

x ∈ (1,+∞ ) x = 2 nula funkcije

x 0

1

2

1 − ln x x2 D: x >0 1 − ln x = 0 ln x = 1

40. y =

y

x ∈ ( 0,+∞ ) x =  nula funkcije y =

1 − ln x x2

y = 1− ln x 41. y = tgx − 3



tgx − 3 = 0 tgx = 3 π x = + kπ nula funkcije 3 y

−π

−

2π 3

π − 2

0

42. Izračunati

f( x) + f 1 

f  1  = log 6

1 9 + 3 log 3 x x

   x

  x

π π 3 2

π 4π

3

3π 2

x

2π

, ako je dat izraz f ( x ) = log 6 x + 3 log3 9 x ?

f  1  = − log 6 x + 3( log 3 9 − log 3 x )    x

f ( x ) + f  1  = log 6 + 3 log 3 9 x − log 6 x + 3( log 3 9 − log 3 x )   x

f ( x ) + f  1  = 3 ⋅ ( log 3 9 + log 3 x ) + 6 − 3 log 3 x = 6 + 3 log 3 x + 6 − 3 log 3 x = 12   x

43. y = − ln x − 2

− ln x − 2 = 0 ln x = −2 nula funkcije 1 x= 2 

22

x

y = −2 − ln x > 0

 1  x ∈  0, 2    

 1  x ∈  2 ,+∞   

y = −2 − ln x < 0 y

y = − ln x − 2 x 1 2

44. y = sin x

x 0 y 0

π 6 1 2

π 4

1

π 3

2 2

3 2

π 2

π

3π 2

2π

1

0

-1

0

sin ( x + 2kπ ) = sin x k = 0,±1,±2,±3... sin x > 0 , x ∈ ( 0 + 2kπ , π + 2kπ )} k ∈ Z sin x < 0 , x ∈ ( π + 2kπ ,2π + 2kπ )

π  π  x ∈  − + 2kπ , + 2kπ  y ↑1−1 2  2  3π π  x ∈  + 2kπ , + 2kπ  y ↓1−1 2 2  y

1

y=sinx x 0

-1

23

π

2π

45. y = arcsin x

x

0

1 2

arcsin x

0

π 6

D : −1≤ x ≤1

2 2

3 2

π 4

π 3

1

0

-1

0

π 2

π

3π 2

2π

y (0,π)

x -1

0

1 y=arcsinx

1 y=sinx -2π

-π

0 -1 y=arcsinx

24

π

2π

46. y = cos x

1 y=cosx -2π

−

3π 2

-π

−

π

0

2

π

3π 2

π

2 -1

x = ( 2k + 1) ⋅

cos x = 0, cos x > 0, cos x < 0,

47. y = arccos x

π 2

,(k ∈ Z )

π  π  x ∈  − + 2kπ , + 2kπ  2  2  3π π  x ∈  + 2kπ , + 2kπ  2 2 

3π 2

x ∈ ( 0, π ) y ↓1−1

π

x ∈ ( π ,2π ) y ↑1−1

2

sin x 48. y = tgx = cos x D:

π 2 sin x = 0 ⇒ x = kπ ( k ∈ Z )

cos x ≠ 0; x ≠ ( 2k + 1) ⋅

tgx = 0

-1

0

1

−

π 2

y=arccosx

−

25

3π 2

2π

y

−

3π 2

−π

−

π 2

0

y=tgx

π

π 2

3π 2

2π

5π 2

y

3π 2

π

π 2

x

0

−

π 2

−π

−

cos x sin x D : sin x ≠ 0 ⇒ x ≠ kπ

3π 2

49. y = ctgx = ctgx = 0

26

cos x = 0 ⇒ x = ( 2k + 1) ⋅

π 2

(k ∈ Z)

x

y

−π

−

π 2

0

y=ctgx

π

π 2

y = arcctgx

2π

3π 2

x

y

2π 3π 2 π

π 2

0 x −

π 2

−π

2x − 3 x+3 D: x+3≠ 0 x ≠ −3 x = −3 vertikalna asimptota y = 0 2x − 3 = 0 -3 3 Znak funkcije x= 2 0 y +

50. y =

2x − 3 x+3

27

+ + -

0

+ + +

y

y=2 2 x

3 2

51. y = shx =

x − − x 2

D=R y = shx = 0

(sinus hiperbolički x)

⇒

x = 0 nula funkcije − x − x x − − x y( − x ) = =− = − y ( x ) funkcija je neparna 2 2 y y=shx x

y

x + − x 52. y = chx = 2 D=R y min ( 0 ) = 1

y=shx

y − x + x x + − x y( − x ) = =− = y ( x ) funkcija je parna 2 2

shx x − − x = 53. y = thx = chx x + − x

x 0

28

x y=1

y=thx y=-1

54. y = cthx =

chx x + − x = shx x − − x

y

y=1 x 0

y=cthx y=-1

Granične vrijednosti funkcije Niz { a n } realnih brojeva ima graničnu vrijednost a ako za proizvoljan ε > 0 postoji prirodan broj n0 takav da je za svaki n > n0 ispunjena apsolutna vrijednost ( a n − n ) ≤ ε označavamo sa

an = a lim n →∞

. Niz koji ima graničnu

vrijednost nazivamo ga konvergetnim, a onaj koji nema vrijednosti nazivamo divergetnim. an =

1 n

0

1 4

1 1 3 2

1

(a-ε, a+ε) - epsilon okoline tačke a

55. a n =

29

5n − 2 n+2

a-ε

(

a

)

a+ε

ε okoline tačke a

an − 5 < ε an − 5 <

ε = 0,1

1 10

5n − 2 1 −5 < n+2 10 5n − 2 − 5n − 10 1 < n+2 10 − 12 1 < n + 2 10 − 120 < n + 2 n > −122 a 0 x n + a1 x n −1 + ⋅ ⋅ ⋅ + a n −1 ⋅ x + a n

lim n n −1 + ⋅⋅⋅ + b n →∞ b x + b x 0

n −1

1

⋅ x + bn

a0 x n + a1 x n −1 + ⋅ ⋅ ⋅ + a n

lim n+ k + bx n + k +1 + ⋅ ⋅ ⋅ + b n →∞ b x 0

lim x →∞ a

−

m n

58.

59.

60.

30

= 0 ,(k ∈ N )

b0 x n + b1 x n −1 + ⋅ ⋅ ⋅ + bn

=∞

1

=

n

am

6

n + 25n

n + 25n / : n

= lim lim 3 x →∞ 8n18 − 4n + 5 x→∞ 3 8n18 − 4n + 5

2

57.

a0 b0

n

a 0 x n+ k + a1 x n + k −1 + ⋅ ⋅ ⋅ + a n+ k

6

56.

=

x +x

= lim lim 5 x →∞ x − 1 x →∞

lim x → −∞ lim x →∞

1 x

3

−

=

1 2

=

1 1 2

6

/ : n6

= lim x →∞

1 1 + 4 3 x x = 0 =0 1 1 1− 5 x =

3

1  3

x −1 x −1 / : x = lim 2 = lim 2 x / : x3 x →∞ x x →∞

1 x3 = 1 = ∞ 1 0 x

1−

25 1 1 n5 =3 = 4 5 8 2 8− 5 + 6 n n 1+

3

(

)

2 2 lim x x + 2 − x − 2 = lim x →∞

= lim x →∞

(

x ⋅ x2 + 2 − x2 + 2

)=

x2 + 2 + x2 − 2 4

x⋅

(

4

)

x −1

lim 4 4 x →1 x ⋅ ( x − 1) ⋅ ( x + 1) ⋅ (

61.

x⋅

(

)(

x2 + 2 − x2 − 2 ⋅

x2 + 2 + x2 − 2

x →∞

4x

lim x →∞

)

x +1

( )

x2 + 2 + x2 − 2

x2 + 2 + x2 − 2 = lim x →1

(

x⋅

)(

(

4 4

=

4 =2 2

x

)(

x +1 ⋅

)=

=

)

x +1

)

1 1 = 1⋅ 2 ⋅ 2 4

1− 3 x 3 x − 1 = 3 x − 13 = 3 x − 1 3 x 2 + 3 x + 1 x −1 1− 3 x 1− 3 x 1 1 1 = = lim − =− =− lim lim 3 2 3 2 1+1+1 3 x →1 x − 1 x →1 3 x − 1 x →1 x + 3 x +1 x + 3 x +1

62. y =

(

63.

)(

)

(1 − sin x )(1 + sin x ) cos 2 x 1 − sin 2 x 1 + sin x = limπ 1 − sin 3 x = limπ 1 − sin 3 x = limπ (1 − sin x ) 1 + sin x + sin 2 x = lim 2 x →0 1 + sin x + sin x x→ x→ x→ 2

2

2

(

)

1+1 2 = 1+1+1 3 x x 1 = sin 2 + cos 2 2 2 (-)  x x 2 x 2 x cos x = cos +  = cos − sin 2 2 2 2 x x x x 1 − cos x = sin 2 + cos 2 − cos 2 + sin 2 2 2 2 2 x 1 − cos x = 2 sin 2 2 x 2 sin 2 1 − cos x 2 =2 = lim 64. lim x →0 sin 2 x x →0 sin 2 x 2 2 3 x 65. y = 2 x −1 D : x2 −1 ≠ 0 =

x2 ≠ 1 x1, 2 ≠ ±1 x3 ( − 1) 3 = − 1 = − 1 = ∞ = Funkcija ima vertikalne asimptote lim lim 2 2 1−1 0 x → −1 x − 1 x →−1 ( − 1) − 1 x3 13 1 1 = = = =∞ lim lim 2 2 1−1 0 x →1 x − 1 x →1 1 − 1 x3 1 1 = lim = =∞ lim 2 nema horizontalne asimptote 0 x →∞ x − 1 x →∞ 1 − 1 2 x x 31

k = lim x →∞

x3 x3 1 x2 −1 = = =1 lim 3 x 1 x →∞ x − x

1 3 3  x3   x3  x −x +x 0 n = lim  2 − kx  = lim  2 − x  = lim = lim x = = 0 2 1 x −1 x →∞  x − 1 x →∞ 1 − 1  x →∞  x − 1  x →∞ 2 x 3 y = x Kosa asimptota funkcije y = x . y x2 −1 x 66. y = 3 − x2 D : 3 − x2 ≠ 0 x2 ≠ 3

x

x1, 2 ≠ ± 3

-1 − 3

x

= lim lim 2 x→− 3 3 − x x →− 3

(

3− − 3

x

= lim lim 2 3 − x x→ 3 x→ 3

3

3−

( 3)

2

=

)

2

=

− 3 − 3 = =∞ 3−3 0

x3 y= 2 x −1

3 3 = =∞ 3−3 0

Ima vertikalnu asimptotu, a kose asimptote nema. x

= lim lim 2 x →∞ 3 − x x →∞

1 x

3 −1 x2

=

0 = 0 Horizontalna asimptota je x – osa. −1 y

− 3

0

x

3

x y= 3 − x2 x2 +1 1− x2 D : 1− x2 ≠ 0

67. y =

32

x 2 ≠ 1 x1, 2 ≠ ±1

1

2

2

x +1 / : x = lim / : x2 x →∞

y = lim lim 2 x →∞ x →∞ 1 − x

1 x 2 = 1 = −1 Horizontalna asimptota y=-1. 1 −1 −1 x2

1+

x2 +1 ( − 1) 2 + 1 2 = lim = =∞ lim 2 2 0 x → −1 1 − x x → −1 1 − ( − 1) x2 +1 12 + 1 2 = = =∞ lim lim 2 2 0 x →1 1 − x x →1 1 − 1

Vertikalne asimptote x= -1 i x=1.

Nema nula funkcije. y

x -1

0

1 y= -1

-1 x= -1

x=1

2

x 3− x D: 3− x ≠ 0 x≠3 x2 32 9 = = = ∞ Vertikalna asimptota je x=3 . lim lim 0 x →3 3 − x x →3 3 − 3

68. y =

x2 x2 1 1 k = lim 3 − x = lim = lim = = −1 2 x −1 x →∞ x →∞ 3 x − x x →∞ 3 − 1 x 2 2  x   x  x 2 + 3x − x 2 3x n = lim  − kx  = lim  + x  = lim = lim = −3 3 − x 3 − x 3 − x x →∞  3 − x x → ∞ x → ∞ x → ∞    y = kx + n y y = − x − 3 Kosa asimptota x 0 -3

y = − x− 3

Horizontalne asimptote nema.

y

-3

0 x

-3

0

-3

33

3

3 x − 1 D: x − 1 ≠ 0

69. y =

y

x ≠ 1 x≠0

x ln x D : ln x ≠ 0 x ≠1

x

x=0

70. y =

x>0

y

ovdje funkcije nema

x 1

x=1 y=f(x)

f ( x +∆y )

∆y

B

- priraštaj funkcije def .

∆ y = f ( xračun Diferencijalni + ∆ x ) − f ( x ) f ( ′x ) =

f ( x)

A( x, f ( x ) ) x

34

- priraštaj argumenta x

∆x

x + ∆x

lim ∆x→ 0

f ( x+ ∆x ) − f ( x ) ∆x

∆y =k ∆x k – koeficijent smjera prave (sekant)

tgα =

B

∆y α

α

A( x, f ( x ) )

∆x

x

∆x → 0 B →A s →t

x + ∆x

t

α – ugao između sekante i x- ose

f ( x+ ∆ x ) − f ( x ) f ( ′x ) = lim = k t A( x, f ( x ) ) ∆x ∆ x→ 0

71. f ( x ) = x + 2 f ( x + ∆x ) = x + ∆x + 2 f ( ′x ) = lim ∆x →0

f ( x + ∆x ) − f ( x ) ∆x

= lim ∆x → 0

x + ∆x + 2 − x − 2 ∆x = lim =1 ∆x ∆x →0 ∆x

x′ = 1

( x + C)′ = 1

72. y = C

f( x) = C f ( x + ∆x ) = C

73. f ( x ) = x

35

f ( ′x ) = lim ∆x →0

C −C 0 = lim =0 ∆x ∆x →0 ∆x

y=C

y′ = 0

f ( x + ∆x ) = x + ∆ x x + ∆x − x x + ∆x + x x + ∆x − x ⋅ = lim = ∆x x + ∆x + x ∆x→0 ∆x ⋅ x + ∆x + x ∆x →0 ∆x 1 = lim = x + ∆x + x 2 x ∆x →0 ∆x ⋅ f ( ′x ) = lim

(

(

)

2

x x −1 ( x + ∆x ) 2 f ( x + ∆x ) = ( x + ∆x ) − 1

74. f ( x ) =

( x + ∆x ) 2 − x 2 ( x + ∆x ) − 1 x − 1 =

f ( ′x ) = lim

∆x

∆x →0

= lim ∆x →0

76. 77. 78.

(x lim ∆x →0

2

)

+ 2 x ⋅ ∆x + ∆x 2 ⋅ ( x − 1) − x 2 ( ( x + ∆x ) − 1) ( ( x + ∆x ) − 1) ⋅ ( x − 1) = ∆x

x 3 + 2 x 2 ∆x + ∆x 2 x − x 2 − 2 x∆x − ∆x 2 − x 3 − x 2 ∆x + x 2 = ∆x ⋅ ( ( x + ∆x ) − 1) ⋅ ( x − 1)

(

)

∆x ⋅ x 2 − 2 x + x∆x − ∆x x 2 − 2x = ( x − 1) 2 ∆x →0 ∆x ⋅ ( x + ∆x − 1)( x − 1) ′ 3x 2 + 2 x − 5 = 6 x + 2 ′  x3  1 2 1 1  − x + arctgx  = 3 x − 1 + = x2 −1+ 2  3  3 1+ x 1+ x2   7 3 ′ 7 4 4 7 4 8 ⋅ x = 8 ⋅ x = 8 ⋅ ⋅ x = 14 ⋅ 4 x 3 4 ′ 1 1  2 1 −5 2 1   x − 2 − 5  = 1+ 3 − ⋅ 6 = 1+ 3 + 6 5 x x 5x  x x x 

= lim

75.

)

(

)

(

)

79.

′ 2 1 1 16 + 2+1 ′ 19 ′ 19 −5 ′ 1 3 2 4   3 12 24   24   24  19 24 −1 19 24 19 = x = x = ⋅x = ⋅x = ⋅  x ⋅ x⋅ x  = x ⋅x ⋅x      24 24 24 24 x 5         f ( ′x ) = lim

f ( x + ∆x ) − f ( x ) ∆x

∆x → 0

f ( x + ∆x ) = ( x + ∆ x ) 3

80. f ( x ) = x 3 f ( ′x ) = lim ∆x →0

= lim ∆x → 0

y=x

36

( x + ∆x ) 3 − x 3

(

∆x 2

y ′ = n ⋅ x n −1

( a) − ( b) = ( 3

= lim ∆x →0

∆x ⋅ 3 x + 3x∆x + ∆x ∆x

n

a−b =

Izvod

3

3

3

3

2

)

x 3 + 3 x 2 ∆x + 3 x ⋅ ∆x 2 + ∆x 3 − x 3 = ∆x

) = 3x

a −3 b ⋅

(

3

2

a 2 + 3 ab + 3 b 2

)

81. f ( x ) = 3 x

f ( x + ∆x ) = 3 x + ∆ x

 3 ( x + ∆x ) 2 + 3 ( x + ∆x ) ⋅ x + 3 x 2  x + ∆x − 3 x  = f ( ′x ) = lim ⋅ 2 3 2 ∆ x  3 ∆x →0  ( x + ∆x ) + 3 ( x + ∆x ) ⋅ x + x    x + ∆x − x ∆x = lim = lim = 2 2  3 3 3 ( x + ∆x ) ⋅ x + 3 x 2  3 ( x + ∆x ) ⋅ x + 3 x 2  ∆x →0 ∆x ⋅  ∆ x → 0 ( ) ( ) x + ∆ x + ∆ x ⋅ x + ∆ x +         1 1 1 = lim = lim = 3 3 3 3 2 2 2 2 3 2 ∆x → 0  3 x2  3 ( x + ∆x ) + 3 ( x + ∆x ) ⋅ x + x  ∆x→0 x + x + x   ∆x − 1 82. f ( x ) = x f ( x + ∆x ) = x + ∆x =1 ∆x  x + ∆x −  x x ⋅ ∆x − 1 f ( ′x ) = lim = lim = x ∆ x ∆ x ∆x →0 ∆x → 0 f ( x + ∆x ) = sin ( x + ∆x ) 83. f ( x ) = sin x 3

(

(

)

)

2 cos

x + ∆x + x x + ∆x − x ⋅ sin 2 2 = ∆x

sin ( x + ∆x ) − sin x = lim ∆x ∆x →0 ∆x →0 2 x + ∆x ∆x cos ⋅ sin 2 2 = cos 2 x = cos x = lim ∆x 2 ∆x →0 2 α +β α −β sin α − sin β = 2 cos ⋅ sin 2 2 ′ f ( x ) = sin x f ( x ) = cos x f ( ′x ) = lim

Izvodi složenih funkcija n

u = n⋅u

n −1

( )′ =  u

u

⋅ u′

⋅ u′

( sin u ) ′ = cos u ⋅ u ′

( cos u ) ′ = − sin u ⋅ u ′

( tgu ) ′ =

( arctgu ) ′ =

′ u

u′ cos 2 u

( ln u ) ′ = 1 ⋅ u ′ = u

(

u

)′

84. x 2 − x + 2 = x 2 − x +2 ⋅ ( 2 x − 1) 3 2 85. (tg x − 3tgx + 3x ) = 3tg x ⋅

′

1

86. (ln ( x − x − 1) )′ = 2 x

−

1 1 − 3⋅ +3 2 cos x cos 2 x 1

2 x −1 x − x −1

37

u′ 1+ u2

87.

′ 1 −1 ′  sin 2 x  2 1 sin 2 x sin 2 x       ′ ⋅  ⋅     1 − sin 2 x  2  1 − sin 2 x  1 − sin 2 x    ln sin 2 x  =  = =  1 − sin 2 x  sin 2 x sin 2 x  1 − sin 2 x 1 − sin 2 x 1 2

′  sin 2 x  ( sin 2 x ) ′ ⋅ (1 − sin 2 x ) − (1 − sin 2 x ) ′ ⋅ sin 2 x  ⋅   (1 − sin 2 x ) 2  1 − sin 2 x  == = sin 2 x sin 2 x sin 2 x 2 2 ⋅ 1 − sin 2 x 1 − sin 2 x 1 − sin 2 x ( cos 2 x ⋅ 2) ⋅ (1 − sin 2 x ) − ( 0 − 2 cos 2 x ) ⋅ sin 2 x 2 cos 2 x − 2 sin 2 x ⋅ cos 2 x + 2 cos 2 x ⋅ sin 2 x (1 − sin 2 x ) 2 = = = sin 2 x 2 sin 2 x(1 − sin 2 x ) 1 − sin 2 x 2 cos 2 x ctg 2 x = = 2 sin 2 x(1 − sin 2 x ) 1 − sin 2 x ′  u  u ′v − uv ′ ′ ′ ′ ′ ′ ′ ′ ′ (c ⋅ u) = c ⋅ u ( u ± v) = u ± v ( u ⋅ v) = u ⋅ v + u ⋅ v   = v2 v  sin 2 x     1 − sin 2 x   =

−

88.

[x ⋅ ln(x + 1 + x ) − 1 + x ]′ = (x ⋅ ln(x + 1 + x ) ′ − ( 1 + x )′ = ′     1 =  x ′ ⋅ ln ( x + 1 + x ) + (ln ( x + 1 + x ) ⋅ x  −  ⋅ 2x  =      2 1+ x  ′      2x  ( x + 1+ x ) = =  ln ( x + 1 + x ) + ⋅ x −  2

2

2

2

2

2

2

2

2

    2 1+ x2  

x + 1+ x2

 

)

(

(

)

   x′ + 1 + x 2 x =  ln x + 1 + x 2 + ⋅ x −     1+ x2 x + 1+ x2   

( = ln ( x +

) )

= ln x + 1 + x 2 +

89.

38

1+ x2

1+ x2 + x 1+ x2 x + 1+ x2

⋅x−

x 1+ x2

)

(

1+

2x

 x 2 1+ x2 ⋅ x −  = ln x + 1 + x 2 + =  2 2 x + 1 + x 1 + x 

(

)

= ln x + 1 + x 2 +

x x + 1+ x2

−

x x + 1+ x2

=

′ ′ ′  2 x a2 x x a x  2 2 2 2      2 a − x + 2 arcsin a  =  2 ⋅ a − x  +  2 arcsin a  =     ′   x   1 +    ′  a2  1 a  = 2 2 2 2   ′  =  x ⋅ a − x + a − x ⋅ x + 2 2  2  x   1 −    a       1  2   a 2  1 x 1 1 a2 − x2 x2 a 2 =  a −x + ⋅ −2 x ⋅   +  ⋅ ⋅ = − + =   2 2 2 2 2 2 2 2 2 a 2  2   2 a −x 2 a −x a −x    x 2 1−     a2 a  

(

( =

a2 − x2

)

2

− x2 + a2

2

2 a −x

 1 90. y = 1 +  

2

)

=

2a 2 − 2 x 2 2

2 a −x

2

=

(

2 a2 − x2 2

2 a −x

)

2

2

= a2 − x2

x

x x

 1  1 ln y = ln1 +  = x ⋅ ln1 +  x x  

′ y′  1    1  ′ = x ⋅ ln1 +  +  ln1 +   ⋅ x y x   x   y′ 1 −1  1 = ln1 +  + ⋅ 2 ⋅x y x  1 x  1 +  x  y′ 1 −1  1 = ln1 +  + ⋅ 1 x y x  1+ x y′ 1  1 = ln1 +  − y x  1  x ⋅ 1 +  x    x   1 1  1  ⋅ 1 +  y ′ = ln1 +  − x x  1    x ⋅ 1 +     x    ln y = x ⋅ ln x ⋅ ( sin x ) 91. y = ( sin x ) x y′ ( sin x ) ′ = ln( sin x ) + x ⋅ cos x = ln ( sin x ) + x ⋅ y sin x sin x

Funkcije

39

A=B ln A = ln B ln x n = n ⋅ ln x

x ⋅ cos x   y ′ = ln ( sin x ) + ⋅ ( sin x ) x  sin x  

92. y = x 3 − 6 x 2 + 9 x − 4 1. Oblast definisanosti D : ∀x ∈ R 2. Parnost funkcije y ( − x ) = ( − x ) 3 − 6 ⋅ ( − x ) 2 + 9( − x ) − 4 = − x 3 − 6 x 2 − 9 x − 4 funkcija niti je parna niti je neparna. 3. Periodičnost funkcije ( ∀ω ≠ 0) funkcija nije periodična y ( x +ω ) ≠ y ( x ) 4. Nule funkcije x3 − 6x 2 + 9x − 4 = 0 x1 = 1

x 3 − 6 x 2 + 9 x − 4 : ( x − 1) = x 2 − 5 x + 4 3

x −x

x 2 − 5x + 4 = 0 x 2 + x3 = 5

2

- 5x 2 + 9x − 4

x 2 ⋅ x3 = 4

- 5x 2 + 5x 4x − 4 4x − 4 ( 0) ( x − 1) ⋅ ( x − 1) ⋅ ( x − 4) = y

x2 = 1 x3 = 4

5. Znak funkcije -∞

1

( x − 1) 2

+

x−4

-

0

-

y

4

+

0

6. Prvi izvod funkcije

(

+∞

y > 0 x ∈ ( 4,+∞ )

+

y < 0 x ∈ ( − ∞ ,1) ∪ (1,4 )

-

0

+

-

0

+

)

y ′ = 3 x 2 − 12 x + 9 = 3 x 2 − 4 x + 3 = 3 ⋅ ( x − 1) ⋅ ( x − 3)

7. Znak prvog izvoda funkcije -∞

3 x −1 x−3

y′ y

8. Grafik funkcije

40

1

+ +

0

3

+∞

+

+

y max (1) = 0

+

+

y min ( 3) = −4

-

-

0

+ +

y

y = x3 − 6x 2 + 9x − 4 x

(1,0) 0

2

1

3

4

-1 -2 -3 -4

93. y = x 3 − 3x + 2 1. Domen D = R = ( − ∞,+∞ ) 2. Parnost y ( − x ) = ( − x ) 3 − 3( − x ) + 2 = − x 3 + 3x + 2 = −( x 3 − 3x − 2) nije ni parna ni neparna. 3. Periodičnost y ( x +ω ) ≠ y ( x )

ω≠0

y ( x +ω ) = ( x + ω ) 3 − 3 ⋅ ( x + ω ) + 2 = x 3 + 3 x 2ω + 3xω 2 + ω 3 − 3 x − 3ω + 2

periodična. 4. Nule x 3 − 3x + 2 = 0 x = 1 pogođena nula

x 3 − 3 x + 2 : ( x − 1) = x 2 + x − 2 x3 − x 2

x2 + x − 2 = 0 x 2 + x 3 = −1

x 2 − 3x + 2

x 2 ⋅ x 3 = −2

x2 − x − 2x + 2 − 2x + 2 ( 0) ( x − 1) ⋅ ( x − 1) ⋅ ( x + 2) = y

x 2 = −2 x3 = 1

-∞

5. Znak( x − 1) 41

x+2 y

2

-2

+

1

+

-

0

-

0

0

+ +

+∞

+ +

0

+

nije

6. Prvi izvod

(

)

y ′ = 3 x 2 − 3 = 3 x 2 − 1 = 3 ⋅ ( x − 1) ⋅ ( x + 1) y′ = 0 x =1 x = −1

7. Znak prvog izvoda -∞

-1

1

+∞

y max ( −1) = 4

x −1

-

x−3

-

+

y′

0

0

+ -

+

y min (1) = 0

+ +

y

8. Grafik

y = x3 − 3x + 2 4

y

3 2 1 x -2

94. y =

-1

0 (1,0)

4x 4 − x2

1.) Domen 4 − x 2 ≠ 0

x ≠ ±2 4 ⋅ ( − x) − 4x  4x  = = −  Neparna funkcija 2.) Parnost y ( − x ) = 2 2 4− x  4 − x2  4 − ( − x)

3.) Periodičnost

42

∀ω ≠ 0

y ( x +ω ) ≠ y ( x )

4.) Nule funkcije y=0

4x = 0 ⇒

x=0

5.) Znak funkcije -∞

-2

-

4x

2− x 2+ x

0

- 0 +

+ + - 0 + + -

y

6.) Prvi izvod

( 4x) ′ ⋅ (4 − x 2 ) − (4 − x 2 ) y′ =

′

(4 − x )

⋅ 4x

2 2

=

(

4x

4 ⋅ ( − 2)

4x

4 ⋅ ( 2)

4x

4 x

= lim lim 2 2 x → −2 4 − x x → −2 4 − ( − 2 ) = lim lim 2 2 x →2 4 − x x → −2 4 − ( 2 )

4 −1 x2

=

=

=

+∞

+

+ 0 + +

)

+ -

4 ⋅ 4 − x 2 + 2x ⋅ 4x

Funkcija uvijek raste y ↑ ( ∀x ∈ D ) 7) Asimptote

= lim lim 2 x →∞ 4 − x x →∞

2

(4 − x )

2 2

=

16 − 4 x 2 + 8 x 2

(4 − x )

2 2

=

4 x 2 + 16

(4 − x )

2 2

−8 =∞ 0

8 =∞ 0

vertikalne asimptote su x= -2 i x= 2.

0 = 0 horizontalna asimptota je y =0. −1 y

Kose asimptote nema. 8) Grafik

y=

4x 4 − x2 x

-2

x= -2 43

> 0 ∀x

-1

0

1

2

x=2

y=0

95. y =

x3

2( x + 1) 2

1.) Domen D : 2( x + 1) 2 ≠ 0

( − x) 3

2.) Parnost y ( − x ) =

2( − x + 1) 2 3.) Periodičnost y ( x +ω ) ≠ y( x )

x ∈ ( − ∞,−1) ∪ ( − 1,+∞ )

x ≠ −1 =

− x3

2( x − 1) 2 ∀ω ≠ 0

funkcija nije ni parna ni neparna.

4.) Nule funkcije x3 = 0

y=0

5.) Znak funkcije

⇒

-∞

x=0 -1

x3

0

-

2( x + 1)

2

+

0

-

y

( x ) ⋅ (2 ⋅ ( x + 1) ) − (2 ⋅ ( x + 1) ) y′ = ′

2

2

-

+

y′ = y′ = y′ =

6 x ( x + 1) − ( 2 ⋅ ( 2 x + 2) ) ⋅ x

(2( x + 1) )

2 2

6 x 4 + 12 x 3 + 6 x 2 − 4 x 4 − 4 x 3

(2( x + 1) )

2 2

3

(

)

(

)

2

(

(

) (

2

)

4

6x x + 2x + 1 − 4x + 4x

= =

(2( x + 1) )

2 2

2 x 4 + 8x 3 + 6 x 2

)

4( x + 1)

4

=

3

)

(

)

2x 2 ⋅ x 2 + 4x + 3 4( x + 1)

4

x 2 ⋅ x 2 + 4x + 3 2( x + 1) 4

y′ = 0

x2 = 0 x1 = 0

x 2 + 4x + 3 = 0 x 2 + x 3 = −4

∨

x 2 ⋅ x3 = 3

( x + 3) ⋅ ( x + 1)

-∞

7.) Znak prvog izvoda

-3

-1

0

+∞

x2

+

+ 0 +

+

x+1 x+3

- 0 + - 0 + + + + 0 +

+ + +

+

+

y′ y

44

3

(

′  3 x 2 ⋅ 2( x + 1) 2 − x 3 ⋅  2 x 2 + 2 x + 1  ⋅x   = 2 2 2 ⋅ ( x + 1)

′

2 2

2

+ +

(2 ⋅ ( x + 1) )

2

0

+

6.) Prvi izvod funkcije 3

+∞

-

+

y max ( −3) = −

27 8

8.) Asimptote

( − 1) 3 = lim lim 2 2 x →−1 2 ⋅ ( x + 1) x → −1 2 ⋅ ( − 1 + 1) x3

x3

=

−1 = ∞ vertikalna asimptota x= -1. 0

1 1 = =∞ 4 2 nema horizontalne asimptote. 0 + 2 + 3 x x x

= lim lim 2 x →∞ 2 ⋅ ( x + 1) x →∞ 2 x3

x3 1 1 2 ⋅ ( x + 1) 2 k = lim = lim 3 = lim = 2 x 2 x →∞ x →∞ 2 x + 4 x + 2 x x →∞ 2 + 4 + 2 2 x x 3 3 3 2   x − x − 2x − 2x  x 1  − 2 x2 + x −1    n = lim  − x  = lim  = = −1 2 2  = lim 2  x →∞  1 2 ⋅ ( x + 1) 2 x →∞  2 ⋅ ( x + 1)  x→∞ 2( x + 1) y = kx + n

(

x3 1 y = x − 1 kosa asimptota funkcije y = . 2 2( x + 1) 2

x

2

0

y

0

-1

9.) Grafik y

y=

-3

-2

-1

0 -1

27    − 3,−  8  

-2 -3 -4

x= -1

45

1

2

)

x3

2 ⋅ ( x +1) 2

x

96. y = x ⋅ − x D : ( − ∞,+∞ ) 1.) Domen ∀x ∈ R 2.) Parnost y ( − x ) = ( − x ) ⋅ −( − x ) = − x ⋅ x funkcija nije ni parna ni neparna. 3.) Periodičnost Nije periodična 4.) Nule funkcije y=0

x=0

5.) Znak funkcije

-∞

0

+∞

x

-

x

+

+

y

-

+

0

+

6.) Prvi izvod funkcije

′ x x x  x  1 ⋅  −  ⋅ x  ⋅ (1 − x ) 1 − x y′ =  x  = = = x 2 x 2 x    y′ = 0 ⇒ 1− x = 0 x =1

7.) Znak prvog izvoda -∞

1

1− x

+

x

+

y′

0

+∞

-

y max (1) =

+

+

0

-

y

8.) Asimptote x

= lim lim x x →∞  x →∞

9.) Grafik

x′

( ) x

′

= lim x ←∞

1 1 = = 0 horizontalna asimptota y = 0. x ∞  y

x 0

1

2

−x

y = x⋅ 46

1 

97. y =

ln x x

1.) Domen D : x > 0 ∧ x ≠ 0 2.) Parnost Funkcija u ( − x ) uopšte nije definisana, nije parna ni neparna. 3.) Periodičnost Nije periodična. 4.) Nule funkcije y=0

ln x = 0 ⇒ x = 1

5.) Znak funkcije

0

1

+∞

ln x

-

x

+

+

y

-

+

0

+

6.) Prvi izvod funkcije 1 ⋅ x − 1 ⋅ ln x 1 − ln x x y′ = = 2 x x2 y′ = 0 1 − ln x = 0 ln x = 1 ⇒

x=

7.) Znak prvog izvoda 0

1− ln x

+∞

+

0

+

y′

-

y max ( ) =

+

+

0

1 

-

y

8.) Asimptote 1 ln x 1 = lim x = lim = 0 horizontalna asimptota je x – osa. lim x →∞ x x →∞ 1 x →∞ x

9.) Grafik

y

ovdje funkcija nema vrijednosti 47

0

1

2

ln x y= x

x

98. y =

x 2 + 3x x+4

x+4≠0

1.) Domen D : 2.) Parnost y( − x )

( − x) 2 + 3 ⋅ ( − x) =

=

−x+4

3.) Periodičnost y ( x +ω ) ≠ y ( x )

x ∈ ( − ∞,−4 ) ∪ ( − 4,+∞ )

x ≠ −4

 − x 2 + 3x  x 2 − 3x  nije parna ni neparna. = −  −x+4 x − 4  

∀ω ≠ 0

4.) Nule funkcije

x 2 + 3x = 0 x( x + 3) = 0 x = 0 ∧ x = −3

5.) Znak funkcije -∞

x

-4

-

-3

0

+∞

- 0 -

x+3 x+ 4

- 0 + - 0 + + y + 6.) Prvi izvod funkcije y′ = y′ =

y > 0 x ∈ ( − 4,− 3) ∪ ( 0,+∞ )

+

y < 0 x ∈ ( − ∞ ,− 4) ∪ ( − 3,0 )

+ + +

( 2 x + 3) ⋅ ( x + 4) − 1 ⋅ ( x 2 + 3x ) = 2 x 2 + 3x + 8 x + 12 − x 2 − 3x ( x + 4) 2 ( x + 4) 2 x 2 + 8 x + 12

( x + 4) 2

7.) Znak prvog izvoda y′ = 0 x1 = −2

x 2 + 8 x + 12 = 0 x 2 = −6

-∞

x+2

( x + 4) 2

x+6 y′

-6

+ +

0 0

-2

+ + -

0

+∞

+ + + +

y max ( −6 ) = −6 y min ( −2 ) = −1

y

8.) Asimptote

x 2 + 3x ( − 4 ) 2 + 3 ⋅ ( − 4 ) 16 − 12 4 = lim = = = ∞ vertikalna asimptota x=- 4. lim −4+4 0 0 x → −4 x + 4 x → −4

k = lim x →∞

48

x 2 + 3x 3 1+ 2 x + 3x x+4 = x = 1 =1 = lim lim 2 x 1 x →∞ x + 4 x x →∞ 1 + 4 x

 x 2 + 3x   x 2 + 3x − x 2 − 4 x  −x −1  = lim n = lim  − x  = lim  = = −1  x+4 x →∞  x + 4  x →∞   x →∞ x + 4 1 y = x − 1 kosa asimptota x

1

0

y

0

-1

9.) Grafik y

y=x-1

-6

-5 -4

-3

-2

-1

( − 2,−1)

0

1 -1 -2 -3 -4 -5 -6 -7

( − 6,−9)

-8 -9

x= -4

49

2

3

4

5

6

x

1

99. y = x ⋅  x 1.) Domen D : 2.) Parnost

x≠0 −

y( − x ) = ( − x ) ⋅ 

1 x

=

3.) Periodičnost

−x 1 x

( ∀ω ≠ 0) nije periodična.

y ( x +ω ) ≠ y ( x )

4.) Nule funkcije 1 x ⋅ x

y=0 ⇒

x = 0 nije iz domena i funkcija nema nula.

=0

5.) Znak funkcije y=

1 x ⋅ x

>0

za

x>0

y=

1 x ⋅ x

<0

za

x<0

6.) Prvi izvod funkcije y′ =

1 ′ x ⋅ x

1 ′ 1 1 1 1 1 1 ′  1 −1 x 1  1 x x x x x x   + x⋅  = 1⋅  + x ⋅  ⋅   =  + x ⋅  ⋅ 2 =  − =  x ⋅ 1 −    x x x  x  

7.) Znak prvog izvoda y′ = 0 x = 1 y min (1) = 

-∞

0

x− 1

1 x

x

y′

1

+∞

+

+

0

+ +

-

+

+

+

-

+

y min ( 1) = 

y

8.) Asimptote 1

1

(

)

1

)

1

x  x ⋅ − 1 ⋅ x −2 1 1 x = y = = =  = =0 lim lim lim lim lim −2 1 − − 1 − − − ∞ − − 1 x x →0 x →0 x →0 x →0 x →0  x x

(

y = lim lim x →0 x →0 +

+

1 x

1 x

= lim x →0 +

1 x

(

1

⋅ − 1 ⋅ x −2 = lim  x = 0 = ∞ = +∞ −2 − 1⋅ x x →0 +

vertikalne asimptota x=0.

50

(

)

)

k = lim

1 x ⋅ x

x

x →∞

= lim

1 x

= 0 = 1

x →∞

′  1    x − 1 1 1 1      1  x  −1  x ⋅ − 1 ⋅ x −2  = n = lim  x ⋅  x − x  = lim x ⋅   x − 1 = lim = lim  = lim −2 ′   x →∞   x →∞ 1 − 1 ⋅ x x →∞  x → ∞ x → ∞ 1        x x

(

=

1 x

= 0 = 1

y = x + 1 kosa asimptota funkcije y=

1 x ⋅ x

.

x

-1

y

0

0 1

9.) Grafik y



(1, ) y=

1 x ⋅ x

1 -3

-2

-1

0 -1

1

2

x

-2 -3

100. y = (3 − x 2 ) ⋅ x 1.) Domen D : 2.) Parnost

(

( ∀x ∈ R ) .

)

(

)

 1  y ( − x ) = 3 − ( − x ) 2 ⋅ − x = 3 − x 2  x  funkcija nije ni parna ni neparna.  

3.) Periodičnost. Funkcija nije periodična. 4.) Nule funkcije -∞ y=0

3 − x2 = 0

5.) Znak funkcije 51

x1, 2 = ± 3

3− x 

+ +

3+x

-

y

+

x

+ + 0

+∞

0

+

+

+

-

+

(

)

)

6.) Prvi izvod funkcije

(

)

(

)

y ′ = −2 x ⋅  x + 3 − x 2 ⋅  x =  x ⋅ − x 2 − 2 x + 3 y′ = 0 x1 = −3

− x 2 − 2x + 3 = 0 ∧ x2 = 1

7.) Znak prvog izvoda

-∞

-3

x+ 3

+

x

1− x y′

1

0

+∞

+ +

+

+

-

+

+ + 0

-

y

8.) Grafik y 5,4

(1,2)

( )

2 x

5

y = 3− x ⋅ 

4 3 2 1 -3

-2 - -1 −6   − 3, 3    

0 -2 -3 -4

2 101. y = ln x − 1

52

x 1

2

3

4

−6 3 y max (1) = 2 y min ( −3) =

1.) Domen D : x 2 − 1 > 0 x ∈ ( − ∞,−1) ∪ (1,+∞ ) . 2 2 2.) Parnost y ( − x ) = ln ( − x ) − 1 = ln x − 1 = y ( x ) Funkcija je parna.

3.) Periodičnost. Funkcija nije periodična. 4.) Nule funkcije x2 −1 = 1

y=0

x2 = 2

5.) Znak funkcije

-∞

x −1 x+1

x− 2

-

-

-1

1

-

+∞

+

- 0 + +

y

x1, 2 = ± 2

+

+ + - 0 + + +

-

-

+

6.) Prvi izvod funkcije y′ =

1 2x ⋅ 2x = 2 x −1 x −1 2

7.) Znak prvog izvoda y′ < 0

x < 0,

y↓

y′ > 0

x > 0,

y↑

8.) Grafik y

y = ln x 2 − 1 -2 - -1

53

0

12

x

x 2 − 2x + 1 x2 +1 1.) Domen D : ( ∀x ∈ R ) .

102. y =

2.) Parnost y ( − x )

( − x ) 2 − 2( − x ) + 1 x 2 + 2 x + 1 = = nije ni parna ni neparna. x2 +1 ( − x) 2

3.) Periodičnost.Nije periodična. 4.) Nule funkcije x 2 − 2x + 1 =0 x2 +1

y=0

( x − 1) 2 = 0

5.) Znak funkcije

0

1

( x − 1) 2

+

6.) Prvi izvod funkcije

0

+

x2 +1

y′ =

x1, 2 = 1 dvostruka nula.

+ +

+

y

+∞

0

+

( 2 x − 2) ⋅ ( x 2 + 1) − 2 x ⋅ ( x 2 − 2 x + 1) = 2 x 3 + 2 x − 2 x 2 − 2 − 2 x 3 + 4 x 2 − 2 x

( x + 1) 2 x − 2 2 ⋅ ( x − 1) 2 ⋅ ( x + 1) ⋅ ( x − 1) y′ = = = ( x + 1) ( x + 1) ( x + 1) 2

2

(x

2

2

2

)

+1

2

2

2

2

2

2

2

7.) Znak prvog izvoda

-∞

x− 1

-

x+1

( x y+′ 1) 2

-1

2

0

1

+

0

+ +

+

+

+

+

-

+

y min (1) = 0 y max ( −1) = 2

y

8.) Asimptote

lim x →∞

2

2

x − 2x + 1 / : x = lim x2 +1 / : x2 x →∞

2 1 + x x2 1 = = 1 horizontalna asimptota y=1. 1 1 1+ 2 x

1−

Kose i vertikalne asimptote ova funkcija nema.

54

9.) Grafik y

5 4

( − 1,2)

x 2 − 2 x +1 y= x 2 +1

3 2

(1,0)

1 -3

-2

-1

0

1

x 2

3

y=1

4

-2 -3 -4

103. y =

3

x +1 x2

1.) Domen D : x ≠ 0 .

( − x) 3 + 1 = − x3 + 1 nije ni parna ni neparna. x2 ( − x) 2

2.) Parnost y ( − x ) =

3.) Periodičnost. Nije periodična. 4.) Nule funkcije y=0 x = −1

x3 + 1 = 0

( x + 1) ( x 2 − x + 1) = 0

5.) Znak funkcije

-∞

x +1

+ + -

2

x − x +1 2

x

y

6.) Prvi izvod funkcije

(

-1

)

0

0

+ + + +

(

0

)

3 x 2 ⋅ x 2 − 2 x ⋅ x 3 + 1 3x 4 − 2 x 4 − 2 x x ⋅ x 3 − 2 x3 − 2 = = = x4 x4 x4 x3 y′ = 0 x3 − 2 = 0 x3 = 2 x=3 2 y′ =

(

)(

x 3 − 3 23 = x − 3 2 x 2 + x ⋅ 3 2 + 3 4

55

)

+∞

+ + + +

7.) Znak prvog izvoda -∞

0

x−3 2

+∞

+ +

x3 y′

0

+ + -

0

y min ( 3 2 ) =

+ + + +

3 3

y

8.) Asimptote k = lim x →∞

x3 + 1 1 1+ 3 3 2 x +1 x x = 1 =1 = lim 3 = lim x 1 1 x x →∞ x →∞

x3 + 1 x3 + 1 − x3 1 1 − x = = lim 2 = = 0 lim 2 2 ∞ x x x →∞ x →∞ x →∞ x y = kx + n y = x kosa asimptota.

n = lim

Horizontalne asimptote nema. x3 + 1 = lim lim x2 x →0 x →∞

9.) Grafik

1 x3 = 1 = 0 vertikalna asimptota je y – osa. 1 ∞ x y

1+

3  3  2, 3  4 

x3 + 1 y= 2 x

1 -3

-2

-1

0 -1 -2

y=x

56

-3

1

2

x

4

≈2

104. y =

2x − 1

( x − 1) 2

1.) Domen D : x − 1 ≠ 0

x ≠1 2 ⋅ ( − x ) − 1 − ( 2 x + 1) = 2.) Parnost y ( − x ) = ( − x − 1) 2 ( x + 1) 2 nije ni parna ni neparna funkcija.

3.) Periodičnost. Nije periodična. 4.) Nule funkcije y=0

⇒

2x −1 = 0

x=

1 2

5.) Znak funkcije -∞

1

2x − 1 ( x − 1) 2

-

y

0

+∞

+

+

+

-

+

+ 0

+ +

6.) Prvi izvod funkcije y′ =

2( x − 1) 2 − 2( x − 1) ⋅ ( 2 x − 1)

( x − 1) 4 2 x ⋅ (1 − x ) y′ = ( x − 1) 4

7.) Znak prvog izvoda

( x − 1) 4

x=0

-∞

0

2x

+

1− x y′ 2x − 1

2x 2 − 4x + 2 − 4x 2 + 2x + 4x − 2

y′ = 0

( x − 1) 4

8.) Asimptote

=

0

∧

x = 1∉ D 1

+∞

+ +

0

+ +

+

+

0

-

-

+

y min ( 0 ) = − 1

-

y 2 −1

1 = +∞ − − 0+ 2x − 1 2 −1 1 = lim = + = +∞ vertikalna asimptota x = 1. lim 2 2 0 x →1+ ( x − 1) x →1+ (1 − 1) 2 1 − 2x − 1 2x − 1 0 x x2 = lim 2 = lim = = 0 horizontalna asimptota. lim 2 1 x →∞ ( x − 1) x →∞ x − 2 x + 1 x →∞ 1 − 2 + 1 x x2 = lim lim 2 2 x →1 ( x − 1) x →1 (1 − 1)

=

Kose asimptote ova funkcija nema. 57

9.) Grafik

y

y=

2x −1

( x − 1) 2

1 2 -3

-2

-1

1

0

( 0,−1)

2

x

-1 -2 -3 x= 1

-4

105.

( x − 2) 3 y= x2 +1

x ∈ ( − ∞,+∞ )

1.) Domen D := R

( − x − 2) 3 − ( x + 2) 3 = = funkcije nije ni parna ni neparna. x2 +1 ( − x) 2 + 1

2.) Parnost y ( − x )

3.) Periodičnost. Nije periodična. 4.) Nule funkcije y=0

x−2=0

x=2 -∞

5.) Znak funkcije

( x − 2) 2 x− 2 6.) Prvi izvod funkcije y′ =

(

)

x2 +1 y

3( x − 2 ) 2 ⋅ x 2 + 1 − 2 x ⋅ ( x − 2) 3

y′ = 0

(x

2

)

+1

x-2 = 0 x1 = 2

2

y′ =

58

x 3 = −3

( x − 2) 2 ⋅ ( x + 3) ⋅ ( x + 1)

(x

2

)

+1

2

+ + -

+∞

0 0

+ + + +

0

( x − 2) 2 [3 x 2 + 3 − 2 x 2 + 4 x ] ( x − 2 ) 2 ( x 2 + 4 x + 3) = =

x 2 + 4x + 3 = 0 x 2 + x 3 = −4 x 2 ⋅ x3 = 3

x 2 = −1

2

(x

2

)

+1

2

(x

2

)

+1

2

7.) Znak prvog izvoda -∞

( x − 2) 2 ( x + 1)

+

-1

+

2

+

0

+∞

+

y max ( −3) =

2

2

x+3 y′

8.) Asimptote

-3

- 0 + - 0 + 0 - 0

+ + + + + + + 0 +

y min ( −1)

( − 5) 3 10

( − 3) 3 = 2

=

− 125 10

=

− 27 2

y prev ( 2 ) = 0

y

( x − 2) 3

3

 2 1 −  3 3 2 ( x − 2) / : x 1 x x + 1 k = lim = lim 3 = lim  = =1 3 x 1 x →∞ x →∞ x + x / : x x →∞ 1 + 1 2 x 3  ( x − 2)   x 3 − 6 x 2 + 12 x − 8 − x 3 − x  − 6 x 2 + 11x − 8  = lim n = lim  2 − x  = lim  =  x2 +1 x2 +1  x +1 x →∞   x →∞  x→∞  11 8 −6+ − 2 −6 x x = lim = = −6 1 1 x →∞ 1+ 2 x ( x − 2) 3 y = x − 6 kosa asimptota funkcije y = . -1 0 6 1 x2 +1 x

Funkcija presijeca kosu asimptotu. y = x−6

( x − 2) 3 y= x2 +1

⇒

( x − 2) 3 − x − 6 = 0

( x − 2) 3 x2 +1

= x−6

x2 +1 x 3 − 6 x 2 + 12 x − 8 − x 3 − x + 6 x 2 + 6 =0 x2 +1 11x − 2 =0 x2 +1 11x − 2 = 0 2 1 x= ≈ 11 5

59

y

-7

-6

0

-5

9.) Grafik y

3 ( x − 2) y=

x2 + 1

( 2,0) -6

-5 -4

-3

-2

-1

0

( − 2,−1)

1 -1 -2 -3 -4 -5 -6 -7

( 0,−8)

-8 -9 -10

− 25    − 3,  2  

-11

-12 -12,5 -13 27  -13,5   − 1,−  2  

106. y = ( x − 1) ⋅ ln 2 ( x − 1) 1.) Domen D : x − 1 > 0 x > 1 x ∈ (1,+∞ ) 2.) Parnost. Nije ni parna ni neparna. 3.) Periodičnost. Nije periodična. 4.) Nule funkcije y = 0 x −1 = 1

60

x=2

2

3

4

5

6 y=x-6

x

5.) Znak funkcije

( ∀x ∈ D )

y>0

6.) Prvi izvod funkcije 2 ⋅ ln ( x − 1) x −1 y ′ = ln ( x − 1) ⋅ [ ln ( x − 1) + 2] y′ = 0 ln  ( x − 1) = 0 ln ( x − 1) + 2 = 0 y ′ = 1 ⋅ ln 2 ( x − 1) + ( x − 1) ⋅

x − 1 = 0

ln  ( x − 1) = −2

x −1 = 1

x - 1 = − 2 x = 1 + −2 1 x = 1+ 2 

x=2

ln ( x − 1)

1

21ln+−x()

7.) Znak prvog izvoda

y′ 8.) Asimptote

= lim x →1+

+

2⋅

-

+∞

-

-

0

+

+

0

-

0

y

+

 1  max  1+ 2    

+ 0

+

y

( x − 1) ⋅ ln 2 ( x − 1) = lim lim x →1 x →1 +

2

x −1 ln ( x − 1) = lim = lim 1 1 x →1+ x →1+ 2 ( x − 1) ln ( x − 1) 2

2 ln ( x − 1) ⋅ − 1⋅

1

1 x −1 =

( x − 1) 2

1 x −1 = 2( x − 1) = 0 + lim 1 x →1+

( x − 1) 2

Vertikalna asimptota funkcije je x= 1. 9.) Grafik

y

5

y = ( x − 1) ⋅ ln 2 ( x − 1)

4 3

 1 4  1+ 2 , 2  ( 1,0)  (2,0) 

2 1 0

61

1

2

3

x 4

=

1 ⋅4 2

107. y = ln

x−3 x+2

x−3 >0 x+2 D : x ∈ ( − ∞,−2 ) ∪ ( 3,+∞ )

1.) Domen ln x > 0

x+2<0

x < −2 x > 3

2.) Parnost. Nije ni parna ni neparna. 3.) Periodičnost. Nije periodična. 4.) Nule funkcije x−3 =1 x+2

⇒ x − 3 = x + 2 0 = 5 funkcija nema nula.

5.) Znak funkcije y>0 y<0

za x < −2 za x > 3

6.) Prvi izvod funkcije

′ ′   x − 3  1  x−3 x + 2 1 ⋅ ( x + 2) − 1 ⋅ ( x − 3) x+2− x+3 y ′ = ln ⋅ ⋅ =  =  = 2 x−3  x+ 2 x−3 ( x − 3) ⋅ ( x + 2) ( x + 2)   x + 2  x+2 5 y′ = ( x − 3) ⋅ ( x + 2) y ′ > 0 ⇒ y raste ( ∀x ∈ D )

7.) Asimptote

x−3

−5 = ln + ∞ = +∞ 0 lim x → −2 x−3 0 ln = lim ln = ln 0 = −∞ lim x + 2 x→3 5 x →3 Vertikalne asimptote su x= -2 i x=3. ln = ln lim x + 2 lim x → −2 x → −2

x−3 x−3 /: x ln = lim ln = ln lim x + 2 x→∞ x + 2 / : x lim x →∞ x →∞

je y=0 , odnosno x- osa. 8.) Grafik

y = ln

3 x = ln 1 = 0 horizontalna asimptota 2 1+ x 1−

y

x−3 x+2

x y=0

62

-2

0

3

108. y = x ⋅ ln 2 x 1.) Domen D : x > 0 x ∈ ( 0,+∞ ) . 2.) Parnost y ( − x ) nije definisano. 3.) Periodičnost. Nije periodična. 4.) Nule funkcije y = 0 ⇒ x = 0 ∉ D ln 2 x = 0 ⇒ x = 1

5.) Znak funkcije

0

1

x

+

ln 2 x

+

0

6.) Prvi izvod funkcije

(

+ +

+

y

+∞

0

+

)

′ 1 y ′ = x ′ ⋅ ln 2 x + x ⋅ ln 2 x = ln 2 x + x ⋅ 2 ln x ⋅ = ln 2 x + 2 ln x = ln x ⋅ ( ln x + 2 ) x y ′ = 0 ln x = 0 ln x + 2 = 0 x =1 ln  x = −2 x = − 2 1 1 x= 2 ≈ 8 

7.) Znak prvog izvoda 0

ln x

ln x + 2 y′

1

+

=

1 1 ⋅ 4 ≈ 2 2

+

y

+

+

y min (1) = 1 ⋅ ln 2 1 = 0

-

+

0

+∞

0

 1  max  2   

y

8.) Asimptote 1 1 2 ln x ⋅ 2⋅ 2 ln x 2 ln x 2x 2 2 x = x = x ⋅ ln x = = = =0 lim lim lim lim lim lim −2 1 1 x x →0 x →0 x →0 1 ⋅ − x x →0 − x x→0 x →0 x x2 x2

(

63

)

9.) Grafik

y

5 4 3 2 1

y = x ⋅ ln 2 x

 1 4  2, 2   ( 1,0)  0

1

x 2

3

4

x=0 2

109. y = sin x − sin x 1.) Domen D : R x ∈ ( − ∞,+∞ ) . 2.) Parnost y ( − x ) = ( sin ( − x ) ) 2 − sin ( − x ) = ( − sin x ) 2 + sin x = sin 2 x + sin x Nije parna ni neparna funkcija. y ( x + 2 kπ ) = ( sin ( x + 2kπ ) ) 2 − sin ( x + 2kπ ) = sin 2 x − sin x 3.) Periodičnost Periodična, sa periodom 2kπ . 4.) Nule funkcije sin 2 x − sin x = 0 sin x ⋅ ( sin x − 1) = 0

sin x = 0

∨

x = kπ

sin x = 1 π x = + 2kπ 2

5.) Znak funkcije

y = sin x ⋅ ( sin x − 1)

0

sin x sin x − 1 y

+ - 0 - 0

+ -

0 0

+

6.) Prvi izvod funkcije

y ′ = 2 sin x ⋅ cos x − cos x = cos x( 2 sin x − 1) 0 y′ = 0 cos x = 0 2 sin x − 1 = 0 π Ekstremi : x1 = π + 2kπ x 2 = +cos 2kxπ 2 6 2 + 1 1 1 1 1 5 y π  =   − = − = − min   - Z0) 2 4 2 4x3 = 2πsin+ x2k−π1 ( k ∈ 2 6 6 y π  = 1−1 = 0 y′

- 0

max   2

2 7.) Znak prvog 1 1  1  izvoda

y

64

y

 5π  min    6 

 3π  max    2 

=  − =− 2 4 2 = ( − 1) 2 − ( − 1) = 1 + 1 = 2

y

+ 0 + + 0 + 0 -

0

-

+ -

+

-

-

8.) Grafik y

2

y = sin x − sin x

 3π   ,2   2 

2

− 2π

−

3π 2

−π

π − 2

0

π   ,0  2 

π

π

 π 1 2  ,−   6 4

 5π 1   ,−  4  6

110. y = cos 2 x + cos x 1.) Domen D : R x ∈ ( − ∞,+∞ ) . 2.) Parnost y ( − x ) = ( cos( − x ) ) 2 + cos( − x ) = cos 2 x + cos x = y ( x ) .

3π 2

2π

3.) Periodičnost. y ( x + 2kπ ) = cos 2 ( x + 2kπ ) + cos( x + 2kπ ) = cos 2 x + cos x = y ( x ) Funkcija je periodična sa periodom 2kπ k = 0,±1,±2... 4.) Nule funkcije y = cos x( cos x + 1) = 0 cos x = 0 π x = ( 2k + 1) ⋅ , 2 (k ∈ Z)

cos x = −1 x = π + 2kπ

0

5.) Znak funkcije

cos x cos x + 1

65

y

+ 0 + + 0 + 0

- 0 +

- 0 -

0

+ + +

x

6.) Prvi izvod funkcije

y ′ = 2 cos x( − sin x ) + ( − sin x ) ⋅ 1 y ′ = − sin x( 2 cos x + 1) y′ = 0 sin x = 0 2 cos x + 1 = 0 x1 = kπ 2cosx = −1 cosx = −

1 2

2π + 2kπ 3 4π x3 = + 2kπ 3 x2 =

7.) Znak prvog izvoda

0

Ekstremi : 2

y

 2π  min    3 

1 1 1 1 1 =  − = − =− 2 4 2 4 2

− sin x

0 - 0 + 2 cos x + 1 + 0 - 0 y′

y max ( π ) = 1 − 1 = 0 2

y

 4π  min    3 

- 0 + 0 -

0

+ + +

y

1 1 1 =  − =− 2 4 2

y max ( 2π ) = ( − 1) 2 − ( − 1) = 1 + 1 = 2

8.) Grafik

y = cos 2 x + cos x

y

( 2π ,2)

( 0,2)

− 111. y = 66

3π 2

sin x 2 + cos x

−π

π   ,0  2 

−

π 2

0

π

3π  4π 1  2  2π 1   3 ,− 4  2  ,−   3 4 π

2π

5π 2

x

2 + cos x ≠ 0 cos x ≠ −2 ( ∀x ∈ R ) . sin ( − x ) − sin x sin x 2.) Parnost y ( − x ) = 2 + cos( − x ) = 2 + cos x = − 2 + cos x = −( y ( x ) ) .Neparna funkcija. 3.) Periodičnost. Periodična sa periodom 2kπ .

1.) Domen D :

4.) Nule funkcije y=0

5.) Znak funkcije

(k ∈ Z)

x = kπ

sin x = 0 0

sin x 2 + cos x

0

+ +

0

y

+

0

+

0

0

-

0

6.) Prvi izvod funkcije y′ =

cos x( 2 + cos x ) + sin x( sin x )

y′ = 0

( 2 + cos x )

2

2cosx + 1 = 0

=

2 cos x + cos 2 x + sin 2 x

( 2 + cos x )

cos x = − x1 =

2

2 cos x + 1

( 2 + cos x ) 2

1 2

2π + 2kπ 3

x2 =

7.) Znak prvog izvoda Ekstremi :

4π + 2kπ 3

0

3 3 3 2 y  2π  = = 2 = 3 max   3  1  3  2 + −  2  2 1 1 − − 2 = 2 =− 1 ≈ 2 y  4π  = min   3 4− 3 4− 3 5  3  2− 2 2 y ′′ =

=

2 cos x + 1

( 2 + cosyx′) 2

+ 0 +

- 0 +

+ +

+

-

+

0

0

y

− 2 sin x( 2 + cos x ) 2 + 2( 2 + cos x ) ⋅ sin x( 2 cos x + 1)

( 2 + cos x ) 4

( 2 + cos x )[ − 4 sin x − 2 sin x cos x + 4 sin x cos x + 2 sin x ] = 2 sin x cos x − 2 sin x = 2 sin x( cos x − 1) ( 2 + cos x ) 4 ( 2 + cos x ) 3 ( 2 + cos x ) 3 y sin x y ′′ = 0 sin x = 0 cos x = 1 y= 2 + cos x x = kπ x=0 x = 2π y ′′ =

Prevojne tačke 8.) Grafik

− 2π 67

−

−

4π 3

−π

 2π 3     3 , 3   

2π 3 0

π 2

2π 3

π

4π 3

 4π 1   ,   3 4− 3

2π

x

112. y = x 2 ⋅ (1 − ln x ) 1.) Domen D : x > 0 x ∈ ( 0,+∞ ) . 2.) Parnost. Nije ni parna ni neparna funkcija. 3.) Periodičnost. Funkcija nije periodična. 4.) Nule funkcije y=0

x2 = 0 ∉ D

∧

ln x = 0 ln  x = 1 x = 1 = 

5.) Znak funkcije

0

+∞

1− ln x

+

x2

+

y

+

6.) Prvi izvod funkcije

0

+

0

-

 1 y ′ = 2 x ⋅ (1 − ln x ) + x 2 ⋅  −  = 2 x(1 − ln x ) − x  x y ′ = x ⋅ ( 2 − 2 ln x − 1) = x ⋅ (1 − 2 ln x ) y′ = 0 ⇒ x = 0 ∉ D ∧ 1 − 2 ln x = 0 2 ln x = 1 1 ln  x = 2 x=

1 2

7.) Znak prvog izvoda

=  0

1− 2 ln x

+

x

+

y′

+

y

68

+∞

0

+

0

-

 y max (  ) = 2

 x  xy−⋅=+20ln1[]() ′ ′  1 xy−=2ln1 ′ xy−=1ln2 ′

− 2 ln x − 1 = 0 1 ln x = − 2

Prevojna tačka

−

x =

1 2

1

=



8.) Asimptote 1 1 − ln x − x −1 ⋅ x 3 x2 2 x x ⋅ (1 − ln x ) = lim = lim = lim = lim =0 lim −3 1 2 x →0 x →0 x →0 − 2 x x →0 x →0 2 x2 −

9.) Grafik

y = x 2 ⋅ ( 1 − ln x )

y

5 4 3

   ,  2 

2 1

x 0 1 1 

113. y =

x +1 x+2

1.) Domen D :

x+2≠0

 2

3

4

x ≠ −2

− x +1

 1− x 1− x = =− 2.) Parnost y ( − x ) = nije ni parna ni neparna funkcija. − x + 2 − ( x − 2) x−2

3.) Periodičnost. Nije periodična. 4.) Nule funkcije. Nema nula. 5.) Znak funkcije

-∞

x+2 y

69

-2

-

0

+∞

+

+

+

-

+

6.) Prvi izvod funkcije y′ =

( )′ x +1

( x + 2) ′

y′ = 0

=

( x + 2) 2

x +1 = 0

x = −1

7.) Znak prvog izvoda -∞ 

( )=

x +1 ⋅( x + 2 ) − 1 ⋅ x +1

-2

(

x +1

≠ 0,

-1

x+1

+ -

+ -

( x + 2) 2

+

+

+

-

+

-

y′

x +1 > 0

)

+∞

x +1

0

( x + 2 − 1) = x+1 ( x + 1) ( x + 2) 2 ( x + 2) 2

x +1

+ +

1 y min ( −1) = = 1 1

y

( x + 2) 2 ⋅ (x+1 ⋅ ( x + 1) + x+1 ) − 2( x + 2) x+1 ⋅ ( x + 1) ( x + 1) 4 ( x + 2) ⋅ (x+1 ) ⋅ [ x 2 + x + 2 x + 2 + x + 2 − 2 x − 2] = x+1 ⋅ ( x 2 + 2 x + 2) y ′′ = ( x + 2) 4 ( x + 2) 3 y ′′ =

y ′′ > 0

Funkcija ima minimum. 8.) Asimptote

lim y = lim x → −∞

x →−∞

x +1 L. P

lim x → +∞ x + 2

=

x +1 L.P x +1 1 1 = lim = lim −( x +1) = = 0 horizontalna asimptota. x + 2 x→−∞ 1 ∞ x →−∞ 

x +1 ∞ = =∞ lim 1 x →+∞ 1

x +1 −1 1 1 = = = = −∞ lim − 0  ⋅ ( − 0) − 0 x → −2 − 0 x + 2 x +1 −1 1 1 = = = = +∞ lim + 0  ⋅ ( + 0) + 0 x → −2 + 0 x + 2

70

9.) Grafik

y

( − 1,1) -3

-2

x +1 y= x+2

1 -1

0 -1

1

2

x

y=0

-2 -3 x= - 2

114. y =

( x − 2) 3

x2 + x +1 1.) Domen D := R

2.) Parnost y ( − x ) =

x ∈ ( − ∞,+∞ )

( − x − 2) 3 = − ( x + 2) 3 funkcije nije ni parna ni neparna. ( − x) 2 − x + 1 x 2 − x + 1

3.) Periodičnost. Nije periodična. 4.) Nule funkcije y=0

x−2=0

x=2

5.) Znak funkcije

-∞

( x − 2) 3 x2 + x +1

6.) Prvi izvod funkcije y′ = y′ =

71

(

)

y

2

+ +

2

( x − 2 ) 2 ( x 2 + 6 x + 5)

(x

2

)

+ x +1

2

)

+ x +1

2

+ +

-

3( x − 2 ) 2 ⋅ x 2 + x + 1 − ( 2 x + 1) ⋅ ( x − 2 ) 3

(x

0

+∞

0 =

+

( x − 2 ) 2 [3 x 2 + 3 x + 3 − 2 x 2 + 4 x − x + 2 ]

(x

2

)

+ x +1

2

y′ = 0

x 2 + 6x + 5 = 0 x 2 + x 3 = −6

x-2 = 0 x1 = 2

x 2 ⋅ x3 = 5 x 2 = −1 y′ =

x 3 = −5

( x − 2) 2 ⋅ ( x + 5) ⋅ ( x + 1)

(x

2

)

+1

2

7.) Znak prvog izvoda

-∞

( x − 2) 2 ( x + x + 1) 2

2

x+5 y′

-5

+

-1

+

- 0 + - 0 + 0 - 0

2

+

+∞

0

+

y max ( −5 ) =

+ + + + + + + 0 +

y min ( −1)

− 343 21

( − 3) 3 = 1

= −27

y prev ( 2 ) = 0

y

8.) Asimptote

( x − 2) 3

3

 2 1 −  3 3 2 ( x − 2) / : x  x = 1 =1 x + x + 1 k = lim = lim 3 = lim 2 3 x 1 x →∞ x →∞ x + x + x / : x x →∞ 1 + 1 + 1 2 x x 3 3 2 3  ( x − 2)   x − 6 x + 12 x − 8 − x − x 2 − x  − 7 x 2 + 11x − 8  = lim n = lim  2 − x  = lim  = 2 2  x + x + 1 x + x + 1 x + x + 1  x →∞  x → ∞ x → ∞     11 8 −7+ − 2 −7 x x = lim = = −7 1 x →∞ 1 + 1 + 1 x x2 ( x − 2) 3 y = x − 7 kosa asimptota funkcije y = . x -1 0 6 1 x2 + x +1

Funkcija presijeca kosu asimptotu.

y = x−7 y=

( x − 2) 3 x2 + x +1

( x − 2) 3

⇒

( x − 2) 3 x2 + x +1

− x−7 = 0 ⇒ x2 + x +1 18 x − 1 =0 18 x − 1 = 0 x2 + x +1

y

= x−7 x 3 − 6 x 2 + 12 x − 8 − x 3 + 6 x + 6 x 2 + 7 =0 x2 +1 1 x= 18

1   ,−7  tačka presjeka kose asimptote i funkcije.  18 

72

-8

-7

-1

-6

9.) Grafik

y

y=

-5

-1

( 2,0)

0

( x − 2) 3 x2 + x + 1

( 7 ,0 )

y=x-7

x

( 0,− 8)  − 343   − 5,  21  

x 115. y = x +1

1.) Domen D : 2.) Parnost y ( − x ) =

( − 1−, 27) x +1 ≠ 0

-27

x ≠ −1

−x

 − x − x = =− nije ni parna ni neparna funkcija. − x + 1 − ( x − 1) x −1

3.) Periodičnost. Nije periodična. 4.) Nule funkcije. Nema nula.

73

-16

-∞

5.) Znak funkcije

-1

x +1

-

x

+

+

-

+

y

6.) Prvi izvod funkcije y′ =

(  )′ x

( x + 1) ′

y′ = 0

=

0

( ) =  ( x + 1 − 1) =

x ⋅( x + 1) − 1 ⋅ x

( x + 1) 2

x ⋅ x = 0

+∞

x=0

x

( x + 1) 2

+

x ⋅ x

( x + 1) 2

x = 0 y =1

7.) Znak prvog izvoda -∞ 

-1

0

+∞

x

x

+ -

( x + 1) 2

+ -

+ 0

0

+ +

+

+

+

-

1 y min ( 0 ) = = 1 1

y

Funkcija ima minimum. 8.) Asimptote

lim y = lim x → −∞

x →−∞

 x L. P x 1 1 = lim = −∞ = ∞ = = 0 horizontalna asimptota. x + 1 x→−∞ 1 ∞ 

 L. P x ∞ = lim = =∞ lim 1 x → +∞ x + 1 x → +∞ 1 x

x −1 1 1 = = = = −∞ lim − 0  ⋅ ( − 0) − 0 x →−1− 0 x + 1 x −1 1 1 = = = = +∞ lim + 0  ⋅ ( + 0) + 0 x →−1+ 0 x + 1

74

9.) Grafik y

x y= x +1 ( 0,1) -3

-2

-1

1 0

1 -1 -2 -3

x= - 1

Integrali

∫ f ( x ) = F( x ) + C F( ′x ) = f ( x )

Osnovni tablični integrali x n+1 +C n +1 ∫ sin xdx = − cos x + C n ∫ x dx =

∫ cos xdx = sin x + C x x ∫  dx =  + C dx ∫ sin 2 x = −ctgx + C dx

∫ cos 2 x = tgx + C x ∫ a dx =

dx

∫ 1− x2

= arcsin x + C

∫ 1+ x2

= arctgx + C

dx

∫ 75

ax +C ln a

dx = ln x + C x

2

x

y=0

116.

∫ (x

3

)

+ 2 x 2 − x − x dx = ∫ x 3 dx + 2∫ x 2 dx − ∫ xdx − ∫ x dx =

x4 x3 x 2 2 x3 +2 − − +C 4 3 2 3

117. x−3= t ( x − 3) 16 dx = = ∫ 3 t 16 dt = ∫ dx = dt

∫3

16 t 3 dt

=

16 +1 t3

19

16 +1 3

( x − 3) 3 +C = 19 3

3 3 ⋅ ( x − 3) 19 + C 19

+C =

118. x −3= t 2 1 x  x  ∫ sin 2 − 3dx = 2 dx = dt = 2∫ sin tdt = 2 ⋅ ( − cos t ) + C = −2 ⋅ cos 2 − 3 + C dx = 2dt

119. dx

∫ ( 2 x − 5) 3

dt 2x − 5 = t 1 1  1  −1 = 2dx = dt = ∫ 23 = ∫ t −3 dt = ⋅  − 2  + C = +C 2 2  2t  t 4 ⋅ ( 2 x − 5) 2 1 dx = dt 2

120.

1 1 1 A B C D Ex + F = 2 2 = 2 = + 2 + + + 2 = 4 2 2 x x x −1 x +1 x +1 x x − 1 x x − 1 x + 1 x ( x − 1)( x + 1) x + 1

)( ) Ax ( x − 1) + B ( x − 1) + Cx ( x + 1) ( x = 2

(

)

4

=

(

4

2

(

(

)

)

(

)

(

)

+ 1 + Dx 2 ( x − 1) x 2 + 1 + ( Ex + F ) x 2 x 2 − 1 = x 2 ( x − 1)( x + 1) x 2 + 1

)(

2

) (

(

)

)(

)

Ax 5 − Ax + Bx 4 − B + Cx 3 + Cx 2 x 2 + 1 + Dx 3 − Dx 2 x 2 + 1 + Ex 5 − Ex 3 + Fx 4 − Fx 2 = x 2 ( x − 1)( x + 1) x 2 + 1

(

)

x 5 ( A + C + D + E ) + x 4 ( B + C − D + F ) + x 3 ( C + D − E ) + x 2 ( C − D − F ) + x ( − A) − B x 2 ( x − 1)( x + 1) x 2 + 1 − B = 1 (1) − A = 0 ( 2) C − D − F = 0 ( 3) C + D − E = 0 ( 4) B + C − D + F = 0 ( 5) A + C + D + E = 0 ( 6) =

A=0 B = −1

(

( 3) − ( 5)

⇒

− B − 2F = 0 - 2 F = -1

76

)

⇒F=

1 2

( 4) − ( 6)

C + D-E = 0 − A−C − D− E = 0 − A − 2E = 0 − 2E = 0 E=0 ( 3) + ( 4) ⇒ 2C − F − E = 0 1 2C = + 0 2 1 C= 4 ( 4) ⇒ D = E −C 1 D=− 4 1 1 1 1 − 2 + − + = 2 4( x − 1) 4( x + 1) 2 x + 1 x = =

⇒

(

)

(

(

)

)

(

)

(

2

2

(

)

4 x ( x − 1)( x + 1) x + 1 2

2

− 4x 4 + 4 + x5 + x3 + x 4 + x 2 − x5 + x 4 − x3 + x 2 + 2x 4 − 2x 2 4 = 2 4 2 4 4x x − 1 4x x − 1

(

)

(

 1 dx 1 1 1 ∫ x 2 x 4 − 1 = ∫  − x 2 + 4( x − 1) − 4( x + 1) + 2 x 2 + 1 1 1 x −1 1 = + ln + arctgx + C x 4 x +1 2 dx 121. ∫ 2 = arctgx + C x +1

(

)

(

)

dx

∫ x2 + x +1 = ∫ 

dx 2

1 3 x +  + 2 4 

4 3 2 3 ⋅ ⋅ arctgt + C = arctg 3 2 3

122. 77

=∫

∫1+

xdx 1+ x

dx 2   1  x +  3  2 ⋅ 4  3     2 

1 2 3 2

x+

3 dt 1 4 = dx = dt = ∫ 22 = 3 t +1  3  2 + 1  3 dt  dx = 2 

1 2 + C = 2 ⋅ arctg  2 x + 1  + C   3 3  3  2

x+

)

 x − 2+1 1 1 1  dx = − + ln x − 1 − ln x + 1 + arctgx + C =  − 2 + 1 4 4 2 

t=

=

)=

− 1 ⋅ 4 ⋅ x − 1 + 1 ⋅ x ⋅ x + 1 ( x + 1) − 1 ⋅ x 2 ⋅ x 2 + 1 ( x − 1) + 2 x 2 x 2 − 1 4

∫1+

xdx

⋅

1− 1+ x

1+ x 1− 1+ x

= ∫ t dt = ∫

1 t 2 dt

=

=∫

1 +1 t2

1 +1 2

=

(

)

(

)

(

)

1+ x = t x ⋅ 1− 1+ x = −1∫ 1 − 1 + x dx = ∫ 1 + x − 1 dx = = dx = dt 1−1+ x 3 t2

3 2

=

2 ⋅ 3

(1 + x ) 3 − x + C

123. 1− x = t 1− x = t2 dx − 2tdt dt  1 ∫ (1 + x ) 1 − x = − dx = 2tdt = ∫ 2 − t 2 t = −2∫ 2 − t 2 = − 2 2 2 dx = −2tdt

(

)

dt

∫

2 −t

+

1 2

∫ 2

1− t2 = x 2

=

2 2

⋅ ln 2 − t −

1 = 2 −t2

2 2 2

1

(

2 −t

)(

2 +t

)

⋅ ln 2 + t = =

A 2 −t

+

1 2

2 −t

ln

B 2 +t

2 +t =

+C =

(

A 2 + At + 2 B − Bt t ( A − B ) + 2 A + 2 B = 2 −t2 2 −t2 A− B = 0 /⋅ 2 2 A + 2B = 1

2 2A = 1

B=

∫

78

1 2 2

=

2 4

2 4

2−x=t − dt = − dx = dt = ∫ = − ln t = − ln 2 − x t 2−x dx = − dt dx

2

) (

2 − 1− x

⋅ ln

)

2 + 1− x

A 2 +t + B 2 −t = 2 −t2

=

A=

1

+C

 = 2 +t dt

124. x +1 = t6

xdx

∫ 3 x +1 − =∫

x +1

= dx = 6t 5 dt = ∫ x = t 6 −1

(

(t 3

)

6

− 1 6t 5 dt

t6 − t6

)

=∫

(

)

(

)

6t 3 ( t − 1) t 5 + t 4 + t 3 + t 2 + t + 1 dt = −6∫ t 8 + t 7 + t 6 + t 5 + t 4 + t 3 dt = − ( t − 1)

= −6

(

)

t9 t8 t7 t6 t5 t4 −6 −6 −6 −6 −6 +C = 9 8 7 6 5 4

− 26 ( x + 1) 9 36 ( x + 1) 8 66 ( x + 1) 7 66 ( x + 1) 5 36 ( x + 1) 4 = − − − ( x + 1) − − = 3 4 7 5 2 2 ( x + 1) 3 − 3 3 ( x + 1) 4 − 6 ( x + 1) 6 x + 1 − x − 1 − 6 6 ( x + 1) 5 − 3 3 ( x + 1) 2 + C =− 3 4 7 5 2

125. x4 1 1 = 1 + = 1 + = x4 −1 x2 −1 x2 +1 ( x − 1)( x + 1) x 2 + 1 = = =

(

(

)

)(

)

(

(

)

)

A( x + 1) x 2 − 1 + B( x − 1) x 2 + 1 + ( Cx + D )( x + 1)( x − 1) = ( x − 1)( x + 1) x 2 + 1

(

A x3 + x + x 2

(

) + 1) + B ( x + x − x − 1) + Cx ( x − 1)( x + 1) ( x + 1) 3

2

3

− Cx + Dx 2 − D

2

)

x 3 ( A + B + C ) + x 2 ( A − B + D ) + x( A + B − C ) + A + B − D = x4 −1 A + B + C = 0 (1) A − B + D = 0 ( 2) A + B − C = 0 ( 3) A − B − D = 1 ( 4)

=

(

(1) − ( 3)

2C = 0

( 2) − ( 4)

D + D = −1

(1) − ( 4)

)

⇒ C=0 D=−

2 B + C + D = −1 1 2 B = −1 + ⇒ 2 1 A+ B +C = 0 ⇒ A− +0 = 0 4

79

=

Ax 3 + Ax + Ax 2 + A + Bx 3 + Bx − Bx 2 − B + Cx 3 − Cx + Dx 2 − D = x4 −1

(

1 2 1 4 1 ⇒ A= 4 B=−

(

)

6t 5 t 6 − 1 dt 6t 5 t 6 − 1 dt 6t 3 t 6 − 1 = = ∫ t 2 (1 − t ) ∫ (1 − t ) = t2 − t3

1 1 1 1 1 1 = 1 + 4 − 4 − 22 = 1 + − − 2 x −1 x +1 x +1 4( x − 1) 4( x + 1) 2 x + 1 dx ∫ x ± a = ln x ± a + C a = const.   x4 1 1 1 1 1 1 ∫ x 4 − 1 dx = ∫ 1 + 4( x − 1) − 4( x + 1) − 2 x 2 + 1 dx = x + 4 ln x − 1 − 4 ln x + 1 − 2 arctgx + C =

(

(

)

)

1 x −1 1 = x + ln − arctgx + C 4 x +1 2

126. x +1

x( x − 1) 3 = = =

(

A B C D A( x − 1) 3 + Bx( x − 1) 2 + Cx ( x − 1) + Dx = + + + = = x ( x − 1) ( x − 1) 2 ( x − 1) 3 x( x − 1) 3

)

(

)

A x 3 − 3x 2 + 3 x − 1 + Bx x 2 − 2 x + 1 + Cx 2 − Cx + Dx x( x − 1)

3

=

Ax 3 − 3 Ax 2 + 3 Ax − A + Bx 3 − 2 Bx 2 + Bx + Cx 2 − Cx + Dx x( x − 1) 3

=

x 3 ( A + B ) + x 2 ( C − 3 A − 2 B ) + x( 3 A + B − C + D ) − A x( x − 1) 3

A+ B = 0 C − 3 A − 2B = 0 3A + B − C + D = 1 − A =1 A = −1 B =1 C +3−2 = 0 C = −1 − 3 +1+1+ D = 1 D=2 A B C D 1 1 1 1 = + + + = − + − + x x − 1 ( x − 1) 2 ( x − 1) 3 x x − 1 ( x − 1) 2 ( x − 1) 3 x +1 1 1 1 dx ∫ x( x − 1) 3 dx = ∫ − x dx + ∫ x − 1 dx − ∫ ( x − 1) 2 dx + ∫ ( x − 1) 3 = dx

=

∫ ( x − 1) 2 dx

∫ ( x − 1) 3

x −1 = t dt t − 2+1 1 1 −2 = = = t dt = =− =− +C  ∫ ∫ 2 − 2 +1 t x −1 t dx = dt  = x −1 = t dt t −3+1 t −2 1 −1 −3 = +C  = ∫ 3 = ∫ t dt = − 3 + 1 = − 2 = − 2 = t t 2( x − 1) 2 dx = dt 

= − ln x + ln x − 1 +

80

1 1 x − 1 2( x − 1) − 1 x −1 2x − 3 − + C = ln + + C = ln + +C 2 2 x − 1 2( x − 1) x x 2( x − 1) 2( x − 1) 2

127.

∫

(1 − x ) 2 dx = x⋅ x

3 − +1 x 2

=

−

3 +1 2

−2

∫

1 − 2x + x 2 3 x2

1 − +1 x 2

−

1 +1 2

  1 2x x 2 dx = ∫  3 − 3 + 3  2 x2 x2 x

1 x2

+

1 +1 2

=

x

−

−

1 2

1 2

− 2⋅

1 x2

1 2

+

3 x2

3 2

 1 1   −3 −   2 2 2 dx = ∫  x − 2 ⋅ x + x dx =     2

+C = −

x

−4 x +

2 3 x +C 3

128.

∫

(x

2

)

(

2

)

1 x2

−1 x x x −1 ⋅ dx = ∫ 2 x x2 3

+1

5 − +1

1 ⋅ x4

dx = ∫ 7

(x

2

)

−1 x2 −

3 x4

dx = ∫

11 x4

− x2

3   11  x 4 x4  dx = ∫  2 − 2 dx = x   x  

3 x4

1

5  3 −  x4 x 4 x4 x 4 4 4 7 4 = ∫  x 4 − x 4 dx = − +C = − = ⋅ x + 4 +C   3 5 7 1 7 x   +1 − +1 − 4 4 4 4 2 x +3 4   129. ∫ 2 dx = ∫ 1 + 2 dx x −1 x −1  4 4 A B A( x + 1) + B ( x − 1) Ax + A + Bx − B x( A + B ) + A − B = = + = = = 2 x − 1 ( x − 1)( x + 1) x − 1 x + 1 x2 −1 x2 −1 x2 −1 A− B = 4 A+ B = 0 2A = 4 A=2 B = −2 x2 + 3 2 2  x −1  ∫ x 2 − 1 dx = ∫ 1 + x − 1 − x + 1 dx = x + 2 ln x − 1 − 2 ln x + 1 + C = x + 2 ln x + 1 + C

130.

∫ =∫

81

 1+ x2  1 − x 2  1+ x2 dx = ∫  + dx = ∫  + 2 2 4   1− x4  1 + x 1 − x 1− x    dx = arcsin x + ln x + x 2 + 1 + C 2 1+ x

1+ x2 + 1− x2 1− x4 dx 1− x

2

+∫

(

)(

)

1− x2 1+ x2 1− x2

(

)(

)

 dx =  

131. x2 +1 = t − x /2

(

)

2t ⋅ 2t − 2 t 2 − 1 t 2 +1 dx = dt x + 1 = t − 2tx + x dx dx 4t 2 2t 2 2 = = = = 1− t ∫ x 2 + 1 ∫ 2t 2 − t 2 + 1 dt = 2t 2 + 2 t2 +1 x2 +1 x = dx = dt = dt 2t 2t 4t 2 2t 2 t = x + x2 +1 2

∫

=∫

(

)

2

2

2t t 2 + 1 dt dt = = ln t + C = ln x + x 2 + 1 + C ∫ 2 2 t 2t t + 1

132.

(

)

x 2 + 2x + 3 = x + t / 2 x 2 + 2 x + 3 = x 2 + 2 xt + t 2 2 x − 2 xt = t 2 − 3 x( 2 − 2t ) = t 2 − 3

− t 2 + 2t − 3

t2 − 3 2(1 − t ) 2 =∫ 2 dt = ∫ x x 2 + 2 x + 3 = x = 2 − 2t t −3  t2 −3  ⋅ +t 2t ( 2 − 2t ) + 2 t 2 − 3 2 − 2t  2 − 2t  dx = dt ( 2 − 2t ) 2 dx

(

dx = dx = dx =

4t − 4t 2 + 2t 2 − 6

( 2 − 2t ) 2

− 2t 2 + 4t − 6

(

( 2 − 2t ) 2

dt

dt

) dt

2 − t 2 + 2t − 3 2 2 (1 − t )

2

− t 2 + 2t − 3

=∫

)

(

)

2 − t 2 + 2t − 3 dt 2(1 − t ) 2 dt = ∫ 2 dt = 2 ∫ = 2 2 2 2 t − 3 t − 3 + 2t − 2t t − 3 − t + 2t − 3 t− 3 t+ 3 ⋅ 2(1 − t ) 2(1 − t )

(

)(

)

(

1   1   2 2 2 3 2 3   = 2∫ − dt = ln t − 3 − ln t + 3 + C = t − 3 t + 3 2 3 2 3     =

82

1 3

⋅ ln

x 2 + 2x + 3 − x − 3 2

x + 2x + 3 − x + 3

+C

)(

)

1 A B At + A 3 + Bt − B 3 t ( A + B ) + A 3 − B 3 = + = = t −3 t − 3 t + 3 t− 3 t+ 3 t− 3 t+ 3

(

2

A+ B = 0

)(

)

(

)(

)

/⋅ 3

A 3 − B 3 =1 2 3A = 1 1 A= 2 3

⇒

B=−

1 2 3

133.

∫ 3

x  5 −  = t 3  −5 2 +1 − 5 3 − dx 1 − 3dt t t 3 9 = − dx = dt = ∫ = −3∫ t 3 dt = −3 ⋅ + C = 3⋅ +C = +C 3 5 5 2 5 2 3 t x x     − +1 23  5 −  5 −  dx = −3dt 3 3 3 3  

134.

∫

dx 3 − 5x 2

x

=

3 5 1 3 5

dx  5x 2 31 − 3 

   

=∫

dx 5x 2 3 1− 3

=∫

dx 3 1−

=

x2 3   5

1

dx

∫ 3

    x   1−  3    5

2

2

=

=t 3 dx = dt =

dx =

83

=∫

3 dt 5

1

∫ 3

5

dt

1− t2

=

1 3

⋅

3 5

⋅∫

dt 1− t2

=

1 5

arcsin t + C =

1 5

arcsin

5x 3

+C

135. x

dx

∫

7 − 11x 2

1

=

7

⋅∫

=∫

dx  11  71 − x 2  7  

7 ⋅ 1−

=

x2  7     11   

1 7

dx

⋅∫

    x   1−  7     11 

2

∫3

=∫

(

(1 − 2 sin

2

)

x + sin 4 x cos xdx 8

( sin x ) 3

8 − +1 t 3

8 − +1 3

)

(

2

)

7 11 dx =

dx = dt =

7 dt 11

2

− 2⋅

2 − +1 t 3

2 − +1 3

+

4 +1 t3

4 +1 3

8 2 4   8 8  −8 − − 2− 4−  3 3 3   = ∫ sin x − 2 ⋅ ( sin x ) 3 + ( sin x ) 3 ⋅ cos xdx = ∫ t − 2t + t 3 dt =        

+C =

t

−

5 3

5 − 3

− 2⋅

1 t3

1 3

+

7 t3

7 3

+C = −

(

3

∫  

∫

2 x − 53 x +

3

5

2  dx dx = ∫ 3 2 x dx − 5∫ 3 x dx + 2 ∫ =  5 4x 4 x  

 2x   3 =t    2 2x x 2 3 3 3 2x  3  dx = ∫  dx = dx = dt  = ∫ t dt = ⋅  3 = 3 2 x 3  2 2 2   3 dx = dt  2  

  3 x = t    1 t 1 t 1 3x 3x t dt ∫  dx = 3dx = dt  = ∫  3 = 3 ∫  dt = 3  = 3   1  dx = dt  3  

3 5

53 sin x

137.

84

=

sin x = t cos 2 x ⋅ cos xdx 1 − sin 2 x ⋅ cos xdx dx = =∫ = = ∫ 8 8 8 cos xdx = dt sin x ( sin x ) 3 ( sin x ) 3

cos 5 x

=

2

=t

7 dt dt 1 1 x 1 11x 11 = 1 ⋅ = arcsin t + C = arcsin +C = arcsin +C ∫ 2 2 11 11 11 7 11 7 1− t 1− t 11

136.

=

dx

=∫

7 11 1

)

− 63 sin x +

33 sin 7 x + C 7

 4x  − 5 = t    4x 4 − dx dx 4 5 5 5 −5x t t   5 ∫ 5 4 x = ∫ 4 x = ∫  dx = − 5 dx = dt  = ∫  ⋅ − 4 dt = − 4 ⋅ ∫  dt = − 4  =   5 dx = − 5 dt  4   3 3 2x 1 3x 5 1 3 ⋅ 3 2 x 5 3 x 5 1 = ⋅  − 5⋅ ⋅ + 2⋅− ⋅ +C = −  − ⋅ +C 5 5 2 3 4 4 x 2 3 2 4 x

138. x +1 =t x x − 1( x + 1) dx = dt x2 1 − 2 dx = dt x 1 2 x +1 dt 2 2 dx = − x dt dx = = ∫ ( t − 1) t  − 2 x  ( t − 1) x + 1 = xt x(1 − t ) = −1 1 x= ( t − 1) 1 x2 = ( t − 1) 2

1 ∫ x2

  = −∫  

1 t 2 dt

1 +1 t2

Parcijalna integracija

( u ⋅ v ) ′ = u ′ ⋅ v + u ⋅ v′ d ( u ⋅ v ) = v ⋅ du + u ⋅ dv

∫ u ⋅ dv = u ⋅ v − ∫ v ⋅ du

′ ∫ d ( u ⋅ v ) = ∫ v ⋅ du + ∫ u ⋅ dv

( u ⋅ v ) − ∫ v ⋅ du = ∫ u ⋅ dv

139.

x=u

∫ x ⋅  dx = dx = du x

x dx = dv v = ∫  dx =  x

x

= x ⋅ x − ∫ x dx = xx − x + C = x ⋅ ( x − 1) + C

140. ln x = u ∫ ln xdx = 1 dx = du x

85

dx = dv x=v

3

2  x + 1 =− =−   +C 1 3  x  +1 2

1 = x ⋅ ln x − ∫ x ⋅ ⋅ dx = x ⋅ ln x − x + C = x ⋅ ( ln x − 1) + C x

2 141. ∫ sin xdx I način :

∫ sin

2

xdx = ∫ sin x ⋅ sin x ⋅ dx =

sin x = u cos xdx = du

sin xdx = dv = I = sin 2 x = - cos x = v

(

)

= − sin x cos x + ∫ cos 2 xdx = − sin x cos x + ∫ 1 − sin 2 x dx = − sin x cos x + x − ∫ sin 2 xdx = I = − sin x cos x + x − I 2 I = x − sin x cos x x − sin x cos x I= +C 2

II način : sin 2 x + cos 2 x = 1 cos 2 x − sin 2 x = cos 2 x 2 sin 2 x = 1 − cos 2 x 1 − cos 2 x sin 2 x = 2 1 − cos 2 x 1 1  2 ∫ sin xdx = ∫ 2 dx = ∫  2 − 2 ⋅ cos 2 x dx =   2 x = t    1 1 1 1 1 1 1 1 = ∫ cos 2 xdx = 2dx = dt  = ∫ cos tdt = sin t = sin 2 x = x − ⋅ sin 2 x + C = x − sin 2 x + C 2 2 2 2 2 2 2 4  1  dx = dt  2  

142.

∫

ln x = u

ln xdx = 1 x2 dx = du x

=−

1 dx = dv 1 1 − ln x x2 = − ⋅ ln x + ∫ 2 dx = + ∫ x − 2 dx = x x 1 x − =v x

ln x x −2+1 − ln x − 1 + +C = +C x − 2 +1 x

143. 2 ∫ x arctgxdx =

86

arctgx = u 1 dx = du 1+ x2

x 2 dx = dv 3

x =v 3

=

x3 1 x3 ⋅ arctgx − ∫ dx = 3 3 1+ x2

  x2 + 1 = t    x3 x  x2 xdx  ∫ 1 + x 2 dx = ∫  x − 1 + x 2 dx = 2 − ∫ x 2 + 1 = 2 xdx = dt  = 3 2  dt  = x arctgx − 1 ⋅  x − 1 ln x 2 + 1  + C    xdx =  3 3 2 2  2   dt x2 x2 1 x2 1 = −∫ 2 = − ln t = − ln x 2 + 1 2 t 2 2 2 2

(

)

144.

dx



x

x

x

∫ sin x = sin x =  2 sin 2 cos 2  = 2tg 2 ⋅ cos

2

x = 2

2tg 1 cos 2

= cos x = cos 2

x x − sin 2 2 2

x 2 = x 2

2tg

x 2

x x sin + cos 2 2 2 x cos 2 2 2

x 1 − tg 2 cos x  2 x 2 x 2 = ⋅ = 1 − tg =  ⋅ cos 1 cos x  2 2 1 + tg 2 x cos 2 2 1 − tg 2

t = tg

=

2tg

x 2

x tg +1 2 2

x 2 = x 2

x 2

x 2dt 2 2 sin x dt x = tgx = = = x = 2arctgt = ∫ 1+t = ∫ = ln t + C = ln tg + C x 2t cos x t 2 1 − tg 2 1 2 2 dx = 2 ⋅ dt 1+t 1+ t2 2t sin x = 1+ t2 x 2tg 2

87

arctgt =

=

145.

cos x =

1− t2 1+ t2

1 − t 2 2dt ⋅ 2 2 1 + t 1 + t 2 =  1 − t =∫ ∫  1 + t 2 1 + t 2 2 1+ t2

x cos xdx 2 ∫ 1 + cos x = 2dt dx = 1+ t2 t = tg

1 + cos x = 1 +

 dt =  

1− t2 1+ t2 +1− t2 2 = = 2 2 1+ t 1+ t 1+ t2

1  x x x   x = arctgt − ∫ 1 − 2 dt = arctgt − t + arctgt + C = 2arctg  tg  − tg + C = 2 ⋅ − tg + C = 2 2 2  t + 1  2 x = x − tg + C 2

146.

x x x x  sin x =  2 sin cos  = 2tg ⋅ cos 2 = 2 2 2 2 

dx 2 x 2 x ∫ 3 + 5 cos x = cos x = cos 2 − sin 2

t = tg

x 2

x 2 x = 2arctgt 1 dx = 2 ⋅ dt 1+t2 2t sin x = 1+t2 1−t2 cos x = 1+t2 arctgt =

88

2tg 1

x 2 =

x cos 2 2

2tg sin 2

x 2

x x + cos 2 2 2 2 x cos 2

cos x  x x ⋅ = 1 − tg 2  ⋅ cos 2 = cos x  2 2

2tg = tg 2

x +1 2

x 1 − tg 2 2 = 1 1 + tg 2 x cos 2 2

1 − tg 2

x 2

x 2 x 2

=

2dt 2dt 2 2dt dt 1+ t2 = ∫ 1+ t 2 = ∫ =∫ = 2∫ = 2 2 2 2 1− t 31+ t + 5⋅ 1− t 3 + 3t + 8 − 5t 8 − 2t 2 3+ 5⋅ 1+ t2 1+ t2 1 4 dt = 1 dt = − 1 du = − 1 ln 2 − t 1   1   2−t 4 ∫ 2−t 4∫ u 4 dt dt 4 4   = 2∫ = = + dt = = ∫ ( 2 − t )( 2 + t ) ∫  2 − t 2 + t  ∫ 2 − t = u  2 4 −t2   − dt = du      dt = −du 

(

(

)

(

)

)

x 1 1 1 2+t 1 2 +C = − ln 2 − t + ln 2 + t = ln = ln x 4 4 4 2−t 4 2 − tg 2 1 A B A( 2 + t ) + B ( 2 − t ) 2 A + At + 2 B + Bt t ( A + B ) + 2 A + 2 B = + = = = ( 2 − t )( 2 + t ) 2 − t 2 + t ( 2 − t )( 2 + t ) ( 2 − t )( 2 + t ) ( 2 − t )( 2 + t ) A+ B = 0 2 A + 2B = 1 2A + 2A = 1 4A = 1 1 1 A= B= 4 4 2 + tg

147. dx

∫ 7 cos 2 x + 3 sin 2 x

dx 1 dx 2 2 : cos x 1 + tg 2 x dx cos x cos x = = = = : cos 2 x ∫ 7 + 3tg 2 x ∫ 7 + 3tg 2 x ∫ 3 + 3tg 2 x 2

(

)

tgx = t x = arctgt dt 1+ t2 dt dt 1 dt dx 1+ t2 = = =∫ =∫ = ∫ = dt ∫ 3 2 2  3  7 7 + 3t 7 + 3t cos x  7 ⋅ 1 + t 2   7 dt   t dx = 1+  2  7 1+ t   3

(

)

7 du 1 1 7 du 7 7 3t 3 = ∫ = ⋅ ⋅∫ = arctgu = arctg +C = 2 2 7 1+ u 7 3 1+ u 7 3 7 3 7 =

89

 3    ( tgx )  + C = 7 arctg  3x  + C arctg  7 3 7 3  7   7  7

t

     

2

=

7 3 dt =

=u = 7 3

du

148. cos x = t 3 2 2 2 ∫ sin xdx = ∫ sin x ⋅ sin xdx = ∫ 1 − cos x ⋅ sin xdx = − sin xdx = dt = − ∫ 1 − t dt = sin xdx = −dt

(

= −t +

)

(

t3 x + C = − cos x + cos 3 + C 3 3

149.

∫ cos

2

)

xdx = ∫ cos x ⋅ cos xdx =

cos x = u − sin xdx = du

(

dv = cos xdx = v = sin x

[∫ cos

2

]

xdx = I =

)

= cos x ⋅ sin x + ∫ sin 2 xdx = sin x cos x + ∫ 1 − cos 2 x dx = sin x cos x + x − ∫ cos 2 xdx = I = sin x ⋅ cos x + x − I 2 I = sin x ⋅ cos x + x sin x ⋅ cos x + x I= +C 2

150.

sin ( α + β ) = sin α cos β + cos α sin β sin ( α − β ) = sin α cos β − cos α sin β sin ( α + β ) − sin ( α − β ) = 2 sin α cos β 1 ∫ sin 3x cos 3xdx = ∫ 2 ( sin 11x + sin ( − 5x ) ) dx =   11x = t    dt 1 −1 cos 11x sin 11xdx = 11dx = dt  = ∫ sin t = ∫ sin tdt = cos t = 11 11 11 11  dt  dx =  11   = =   − 5 x = t    1 1 1  dt  ∫ sin ( − 5x ) dx = − 5dx = dt  = ∫ sin t ⋅  − 5  = − 5 ∫ sin tdt = 5 cos t = 5 cos( − 5x )  dt  dx =  −5   =

1  − cos11x cos 5 x  +  +C 2  11 5 

151.

∫ sin 3x sin ( 5x − 1) dx

cos( α + β ) = cos α cos β − sin α sin β cos( α − β ) = cos α cos β + sin α sin β

cos( α − β ) − cos( α + β ) = 2 sin α sin β sin α sin β =

90

1 ⋅ [ cos( α − β ) − cos( α + β ) ] 2

1

1

∫ sin 3x sin( 5 x − 1) dx = 2 ∫ [ cos( 3x − 5 x + 1) − cos( 3x + 5x − 1) ]dx = 2 ∫ [ cos(1 − 2 x ) − cos( 8 x − 1) ]dx =   1 − 2 x = t    1 1 1  dt  ∫ cos(1 − 2 x ) dx = − 2dx = dt  = ∫ cos t ⋅  − 2  = − 2 ∫ cos tdt = − 2 sin t = − 2 sin (1 − 2 x )  dt  dx = −  2   = =   8 x − 1 = t    dt 1 1 1 ∫ cos( 8x − 1) dx = 8dx = dt  = ∫ cos t 8 = 8 ∫ cos tdt = 8 sin t = 8 sin ( 8x − 1)  dt  dx =  8   =

1 1 1 1 1  − sin (1 − 2 x ) − sin ( 8 x − 1)  + C = − sin (1 − 2 x ) − sin ( 8 x − 1) + C  2 2 8 4 16 

152. cos x = t dx sin xdx sin xdx dt ∫ sin x cos 2 x = ∫ sin 2 x cos 2 x = ∫ 1 − cos 2 x cos 2 x = − sin xdx = dt = −∫ 1 − t 2 t 2 = sin xdx = −dt

(

)

(

1 1 A B Ct + D = = + + = 2 2 2 1−t t (1 − t )(1 + t )t 1 − t 1 + t t2

(

)

(

(

)

(

)

=

A(1 + t ) ⋅ t 2 + B 1 − t 2 ⋅ t 2 + ( Ct + D ) 1 − t 2 ( A + At ) t 2 + ( B − Bt )t 2 + Ct − Ct 2 + D − Dt (1 + t ) = = (1 − t )(1 + t ) t 2 (1 − t )(1 + t )t 2

=

At 2 + At 3 + Bt 2 − Bt 3 + Ct + Ct 2 − Ct 2 − Ct 3 + D + Dt − Dt − Dt 2 = (1 − t )(1 + t )t 2

=

t 3 ( A − B − C ) + t 2 ( A + B − D ) + Ct + D (1 − t )(1 + t ) t 2

= A − B −C = 0 A+B−D =0 C =0 D =1 1 1 1 = 2 + 2 + 2 1−t 1+ t t

= −∫

91

)

)

dt 1−t2 t2

(

)

2A − C − D = 0

A=

1 2

B=

1 2

2 A −1 = 0

1  1     1 1 1 t −2+1  2 2 +C = = −∫  + + 2 dt = − − ln 1 − t + ln 1 + t +  2 2 − 2 + 1 1 − t 1 + t t       

=

=

1 1 1 1 1 − cos x 1 1 − cos x 1 ln 1 − cos x − ln 1 + cos x + + C = ln + + C = ln + +C 2 2 cos x 2 1 + cos x cos x 1 + cox cos x

x y ln x + ln y = ln x ⋅ y ln x − ln y = ln

1

∫ cos nxdx = n sin nx + C

1

1 ln x = ln x 2 = ln x 2 ln x n = n ⋅ ln x

153.

2

(

)

1 + cos 2 x 1  1 + cos 2 x  2 ∫ cos xdx = cos x = 2 = ∫  2  dx = 4 ∫ 1 + 2 cos 2 x + ( cos 2 x ) dx = 4

2

  2 x = t    1 + cos 4 x 1  1 + cos 4 x  dt 1 cos 2 2 x = = ∫ 1 + 2 cos 2 x + dx = ∫ cos 2 xdx = 2dx = dt  = ∫ cos t = sin t =  2 4  2 2 2   dt  dx =  2   =

1  1 1 1 1 3 1 1  ⋅  x + 2 ⋅ sin 2 x + x + ⋅ sin 4 x  + C = x + sin 2 x + sin 4 x + C 4  2 2 2 4 8 4 32 

154.

∫x

dx x2 −1

=

1 =t x 1 x= t dx = −

1

dt t −2 1 1− t2 x2 −1 = 2 −1 = t t2

(

= − arcsin

)

− dt − dt 2 dt t t2 =∫ =∫ = −∫ = − arcsin t + C 1 1− t2 1 1− t2 1− t2 ⋅ ⋅ t t t t2

1 +C x

155. 1 2t + 1 x = 2+ = t t 2 1 4t + 4t + 1 − dt t= x2 = 2 x−2 t dx t2 =∫ = ∫ x − 2 x 2 − 6x + 1 = 2 2 − dt 4t + 4t + 1 12t + 6 2 1 1 − 2 t − 7 t dx = 2 x − 6x + 1 = − +1 = ⋅ t t t2 t t2 2 2 2 2 4t + 4t + 1 − 12t + 6t + t 1 − 2t − 7t = = 2 t t2 x−2=

92

1 t

dt dt dt dt t2 = −∫ = −∫ = −∫ = −∫ =  2 2  1 1 − 2t − 7t 2 1 − 2t − 7t 2 1 − 2t + 7t 2 1 − 7 t + t  ⋅ 7  t t  dt dt dt dt = −∫ = −∫ = −∫ = −∫ 2 2  1  2 1    1 2 1 8  1  1  t +  1 − 7 t +  + − 7 t +  1 − 7 ⋅  t +  −  7 7 8  7 49   7  7  7  1− 8 7   49 

(

1

=−

⋅∫

8 7

7

=−

8 ⋅ 7 ∫

⋅

8

 1 t + dt = 7 =u 2  8  1  7  t +   7   1−  8     7 

 arcsin 7 

1

=−

dt

)

= du

8 7

      

=

 8  dt = du  = 7  

   1 1 t + 1   +  du 1 7  + C = − 1 arcsin x − 2 7  + C = =− arcsin    7 8  7 8 1− u2     7  7    7+ x−2  1 x+5 +C = − arcsin +C 8 ⋅ ( x − 2 )  7 8 ⋅ ( x − 2)

156.

∫ x⋅(

dx x + 5 x2

x = t 10

10t 9 dt

) = dx = 10t dt = ∫ t ( t 9

10

10

+ 5 t 20

)

10t 9 dt 10t 9 dt dt = ∫ 10 5 4 = ∫ 9 = 10 ∫ 5 = 4 t t +t t ⋅ t ⋅ t ( t + 1) t ( t + 1)

(

)

1 At 4 + Bt 3 + Ct 2 + Dt + E F = + = 5 5 ( t + 1) t ( t + 1) t =

( At

4

)

+ Bt 3 + Ct 2 + Dt + E ⋅ ( t + 1) + t 5 F At 5 + At 4 + Bt 4 + Bt 3 + Ct 3 + Ct 2 + Dt 2 + Dt + Et + Ft 5 = = t 5 ( t + 1) t 5 ( t + 1)

t 5 ( A + F ) + t 4 ( A + B) + t 3 ( B + C ) + t 2 ( C + D) + t( D + E ) + E = = t 5 ( t + 1) E =1 D+E =0 C+D=0 B+C =0 A+ B = 0 A+ F = 0 D = −1 C =1 B = −1 A =1 F = −1 =

 t4 − t3 + t2 − t +1 1  1  dt = 10 ∫  t −1 − t − 2 + t −3 − t − 4 + t −5 − = 10∫  − dt = 5  t + 1 t + 1 t    

93

t −2+1 t −3+1 t −4+1 t −5+1 = 10 ln t − 10 + 10 + 10 − 10 − ln t + 1 + C = − 2 +1 − 3 +1 − 4 +1 − 5 +1 10  x 1 1 1 1  = 10ln 10 + 10 − 5 + − +C x +1 x 2 x 310 x 3 45 x 2  

157. x = 3 sin t dx = 3 cos tdt dx 3 cos tdt cos tdt cos tdt x =∫ = 3∫ = 3∫ = ∫ 3 − x 2 = sin t = 3 3 − 3 sin 2 t 3 1 − sin 2 t 3 1 − sin 2 t x t = arcsin 3 cos tdt x =∫ = ∫ dt = t = arcsin +C 3 cos 2 t

(

)

158. 1 ( 2 sin x cos x ) 2 cos 2 xdx = 4 1 1 1  sin 2 x sin 2 2 x ⋅ cos 2 x   1 + cos 2 x  1 1  2 dx = = ∫ ( sin 2 x ) 2 ⋅  + dx = ∫ sin 2 x ⋅  + cos 2 x dx = ∫   4 2 4 4  2 2   2 2   1 1 1 − cos 4 x 1 = ∫ sin 2 2 x + sin 2 x cos 2 x dx = ∫ dx + ∫ sin 2 2 x cos 2 xdx = 8 8 2 8 3 1 1 1 1  1 sin 2 x 2 = ∫ (1 − cos 4 x ) dx + ∫ sin 2 x cos 2 xdx =  x − sin 4 x  + ⋅ +C 16 8 16  4 6  8

∫ sin

2

x cos 4 xdx = ∫ ( sin x cos x ) 2 cos 2 xdx = ∫

(

)

  4 x = t    1 1 ∫ cos 4 xdx = 4dx = dt  = 4 ∫ cos tdt = 4 sin 4 x  dt  dx =   4  sin 2 x = t  1 2 1 1 t 3 t 3 sin 3 2 2 sin 2 x cos 2 xdx = = t dt = ⋅ = = 2 cos 2 xdx = dt  2 ∫ 2∫ 2 3 6 6   =

x sin 4 x sin 3 2 x − + +C 16 64 48

Određeni integrali 94

Integralna suma za funkciju y = f ( x ) na intervalu [ a, b] i podjeli n

a = x0 < x1 < ⋅ ⋅ ⋅ < x n = b tog intervala je

∑f (ξ )∆xi

.

i =1

xi −1 ≤ ξ i ≤ xi , ∆xi = xi − xi −1 , i = 1,2,3,..., n b

n

lim ∑ f ( ξ i )∆xi = ∫ f ( x ) dx = Px

max ∆xi →0 i =1

a

Γf f ( ξ2 ) b

∫ f ( x ) dx

a

a

b

Osobine određenih integrala a

∫ f ( x ) dx = 0 a

ako je funkcija neparna. ako je funkcija parna funkcija. Newton-Leibnitz-ova formula

Funkcija f ( x ) neprekidna na [ a, b] i ϕ ( x ) njena primitivna funkcija tj. ϕ ( x ) = f ( x ) b

tada je

∫ f ( x ) dx = ϕ ( x ) a

159. 95

b a

= ϕ( b) − ϕ( a) .

4

x2 xdx = ∫ 2 0

4

42 02 − =8 2 2

= 0

160. 1

x n +1 ∫ x dx = n + 1 0

1n +1 0 n +1 1 = − = n +1 n +1 n +1

1

n

0

161. 0

3x 4 + 3x 2 + 1 dx = x2 +1

∫π

−

3x 4 + 3x 2 + 1 1  x3  2 3 ∫ x 2 + 1 dx = ∫  3x + x 2 + 1 dx = 3 3 + arctgx = x + arctgx =

4

(

= x 3 + arctgx

)

0 −

π 4

3

π  π  π π = 0 + 0 −  −  + arctg  −  = + arctg 4  4  4  64 3

162.

A B A( x − 1) + B( x + 2 ) Ax − A + Bx + 2 B + = = = ( x + 2)( x − 1) ( x + 2)( x − 1) x + x − 2 x + 2 x −1 x( A + B ) − A + 2 B = A+ B = 0 ( x + 2)( x − 1) +∞ dx − A + 2B = 1 = ∫ x2 + x − 2 = 2 3B = 1 1 1 B= A=− 3 3 1

2



=

1 1      3 + 3 dx =  − 1 ln x + 2 + 1 ln x − 1  =  ln 3 x − 1   3 x + 2   x + 2 x −1  3      

+∞ 

∫ 2

=

−

+∞

=

2

2

− 1 2 = − ln 3 = − ln 2 3 = ln 2 4 3

163. y = 4x ⇒ y = 2 x 4 4 y = x+ 5 5 Presjek funkcija f1 i f2 y 2 = 4x 4 4 y = x+ 5 5 2

4 4  x +  = 4x 5 5

96

( f1 ) ∩ ( f 2 ) je rješenje sistema jednačina.

2

(

)

4 2   x + 1 = 4x 5 16 2 25 x + 2x + 1 = 4x ⋅ 25 4 2 4 x + 8 x + 4 = 25 x

(

)

4 x 2 − 17 x + 4 = 0 17 ± 289 − 64 8 17 ± 225 17 ± 15 x1, 2 = = 8 8 2 1 x1 = = 8 4 32 x2 = =4 8 x1, 2 =

3     4 2 2 4x 4  x 4 x 4  4 3 2 2 4   P = ∫2 x − − dx = 2 ⋅ − ⋅ − x = x − x − x 3 5 2 5  1 3 5 5 5 5   1   4 4 2   4 2 4  4 1 2 1 4 1   32 32 16 1 1 1 =  ⋅ 8 − ⋅ 16 − ⋅ 4 −  ⋅ − ⋅ − ⋅   = − − − + + = 5 5 5 5 6 40 5  3 8 5 16 5 4   3 3 4

=

1280 − 768 − 384 − 20 + 3 + 24 135 27 9 = = = 120 120 24 8

164. y 2 = 2x + 1 x − y −1 = 0

( 4,3)

x = y +1 y 2 = 2( y + 1) + 1 y2 = 2y + 2 +1 y2 − 2y − 3 = 0 y1 = −1 y2 = 3 x = y +1 x1 = −1 + 1 = 0 x2 = 3 + 1 = 4

97

 1   − ,0   2 

P∆

4 1 4

=

0

P=2∫

1 − 2

 = 2  

1

4

1 2 x + 1dx + + ∫ 2 x + 1dx + ∫ 2 0 1

( 2 x + 1) 3   

3

0 1 − 2

+

1  + 2  

(

)

2 x + 1 − ( x − 1) dx =

( 2 x + 1) 3   

3

 +  0 

1

( 2 x + 1) 3   

3

 x2 −  1  2 4

   

4

+x

1

4

=

1

1 2 1 1 1 16 1  1  3 3 1 27 3 3   = 2 ⋅  − 0  + +  − + − − 8 − + 4 − 1 = + − + 1 + + 3 =   3 3 3   2 3 2 3 2 3 3  2  3

165. x2 2 1 y= 1+ x2 x2 = 2y 1 y= 1+ 2y y=

2y2 + y −1 = 0

y

y=

x2 2

(1,0)

y=

1 1+ x2

0

−1± 1+ 8 −1± 3 = 4 4 y1 = −1 y1, 2 =

x

1 2 = −2 ∈ R

y2 = x1, 2

x3, 4 = ±1 1  1 x2 P = ∫  − 2 2 −1 1 + x π 1 3π − 2 = − = 2 3 6

 1 x3 dx = arctgx − ⋅  2 3 

1 −1

= arctg1 −

1  1 1 −  arctg ( − 1) +  = 2arctg1 − = 6  6 3

166. Odrediti zapreminu V koja nastaje rotacijom prave y = y

y=

1 1+ x2 y=0

x= − 1 98

x=1

1 oko x – ose. 1+ x2

1+ x2 − x2

1+ x2 x 2 dx dx − ∫ 1+ x2 2 ∫ 1+ x2 2 = 1+ x2 dx x xdx =∫ − ∫x⋅ dx = arctgx − ∫ x ⋅ 2 2 1+ x 1+ x 1+ x2

(

1

2

1

 1  V = π ∫ y 2 dx = π ∫  dx = 2  1 + x   −1 −1

)

dx = ∫

(

(

)

(

)

)

(

)

2

=

x 1 1 x = arctgx + − arctgx = arctgx + 2 2 2 2 1+ x 2 1+ x2 xdx ∫ x ⋅ 1 + x 2 2 =I1

(

(

)

(

=

)

)

  xdx ′ x=u = v   1+ x2       2 1+ x = t   xdx  I1 = x ⋅ xdx = dx = du v=∫ = 2 xdx = dt = ∫ 2 2 2 2   1 + x 1 + x = dt = xdx =   2   1 dt 1 t − 2+1 1 1   v= ∫ 2 = ⋅ =− =− 2   2 t 2 − 2 +1 2t 2 1+ x   −x 1 x 1   +∫ dx = − + arctgx =  2 2 2 2 2 1+ x 21+ x  2 1+ x  1  1  1 x 1  1 1  1  = π ⋅  arctgx + = π ⋅  arctg1 +  −  arctg ( − 1) −  = π  arctg1 +  = 2  2⋅2  2 2 ⋅ 2  2 2 1 + x  −1   2 2

(

(

)

)

(

)

(

(

)

(

(

)

(

)

)

2 π π 2 + 2π π 1 π = π ⋅ +  = + = 2 4  4 2 4

Diferencijalne jednačine 99

)

Diferencijalne jednačine prvog reda y = y ( x ) , y ′ = f ( x, y ) . Oblik jednačine je y ′ = f ( x, y ) Tipovi diferencijalnih jednačina 1.) Jednačina koja razdvaja promjenljive f ( x ) dx = F( y ) dy rješenje : ∫ f ( x ) dx = ∫ F( y ) dy + C . 2.) Homogena diferencijalna jednačina y ′ = ϕ y    x

 y x

pomoću smjene   = z

y = x⋅z

y ′ = z + x ⋅ z ′ rješenje jednačine.

3.)Linearna diferencijalna jednačina y′ + f ( x) y = g ( x) f ( x ) , g ( x ) - neprekidne funkcije − f ( x ) dx  f dx ⋅ ∫ ∫ ( x ) ⋅ g ( x ) dx + C  rješenje : y =  ∫  



 dx

ako je linearna diferencijalna jednačina data u obliku dy + f ( y ) ⋅ x = g ( y ) − f ( y ) dy  f dy ⋅ ∫ ∫ ( y ) ⋅ g ( y ) dy + C  . tada je rješenje : x =  ∫  





3.) Bernulijeva diferencijalna jednačina y′ + f( x) y = g ( x) ⋅ y n y′ z′ = yn 1− n ydx − xdy = 0 ydx = xdy dx dy = x y ∫ dx dy ∫ x =∫ y ln x = ln y + C ln

(

100

)

xy + x ⋅ y ′ = y x

(

)

y +1 ⋅

)

y′ =

dy =y dx

y + 1 dy dx = y x

1 y

n −1

= z z ′ = (1 − n )

z′ + f ( x ) z ′ = g ( x ) linearna diferencijalna jednačina 1− n

x =C y

167.

(

( n ≠ 0 ∧ n ≠ 1) uvodi se smjena

∫

dy dx

y′ yn

∫

(

)

y + 1 dy dx =∫ y x

1  1 − 1   ∫  y + y dy = ∫ x 2 dx   1  −1+1   y2  x − 2 +1  + ln y  = +C  − 1 +1  − 1 +1   2  2  1

y2 + ln y = 2 x + C 1 2 2 y + ln y = 2 x + C

168.

169.

y ⋅ y′ + x = 1 dy y⋅ + x =1 dx dy y⋅ = 1− x dx

y ⋅ dy = (1 − x ) dx

(

dy = x dx x  dx ydy = 2 1 + x ∫

(

)

x

 dx

∫

∫ ydy = ∫ 2(1 + x )

∫ ydy = ∫ (1 − x ) dx 2

)

2 ⋅ 1 + x y ⋅

y2 1 = ⋅ ln 1 + x + C ⋅ 2 2 2

2

y x = x− +C ⋅2 2 2 y 2 = 2x − x 2 + C

y 2 = ln 1 + x + C

170.

y − xy ′ − yy ′ = x y − x = y ′( x + y ) y−x = y′ y+x y−x :x y′ = y+x :x y −1 y′ = x y +1 x z −1 z + xz ′ = z +1 dz z − 1 z+x = ⋅ ( z + 1) dx z + 1 ( z + 1)( z + xz ′) = z − 1

101

y =x x

y = xz

y ′ = z + xz ′

z 2 + xz ′ ⋅ z + z + x ⋅ z ′ − z = −1 z ′x( z + 1) + z 2 = −1 −1− z2 1+ z dz − 1 − z 2 x⋅ = dx z +1 ( z + 1) dz = dx x −1− z2 ( z + 1) dz = dx − x ∫ 1+ z2 ( z + 1) dz = dx −∫ ∫x 1+ z2 1  −  ln 1 + z 2 + arctgz  = ln x + C 2  xz ′ =

1 y2  y  −  ln 1 + 2 + arctg    = ln x + C 2 x  x    1 y2  y − ln 1 + 2 − arctg   = ln x + C 2 x  x y 1 x2 + y2 − arctg = ln x ⋅ C + ln x 2 x2 − arctg

y x2 + y2 = ln C ⋅ x + ln x x2

− arctg

y = ln C ⋅ x 2 + y 2 x

171.

x y′ = y x +1 x( x + 1) = y ⋅ y ′ dy x( x + 1) = y dx x ⋅ ( x + 1) dx = ydy

∫ (x

2

)

+ x dx = ∫ ydy

x3 x2 y 2 + = + C ⋅6 3 2 2 2 x 3 + 3x 2 = 3 y 2 + C

102

  1 + z 2 = t    1 dt 1 z ∫ 1 + z 2 dz = 2 zdz = dt  = 2 ∫ t = 2 ln z  dt   zdz =   2  1 = ln 1 + z 2 2

172.

7 x + xy + y ′( y + xy ) = 0 7 − −1 dy y = 1 dx +1 x dy dx = 7 1 − −1 +1 y x dy dx = − 7 − y 1+ x y x ydy xdx = − 7 − y 1+ x ∫ ydy xdx ∫ − 7 − y = ∫1+ x y+7−7 x +1−1 −∫ dy = ∫ dx 7+ y 1+ x  y+7  7 dx −  ∫ dy + ∫ dy  = ∫ dx − ∫ 7+ y  1+ x  7+ y − y + 7 ln 7 + y = x − ln 1 + x + C

174. x 2 + y 2 = 2 xyy ′ y′ =

x2 + y2 / : x2 2 xy / : x 2

y2 1+ 2 x ′ y = y 2 x 1+ z2 z + xz ′ = 2z 2 1+ z xz ′ = −z 2z 1 + z 2 − 2z 2 xz ′ = 2z dz 1 − z 2 x = dx 2z

103

y =z x

y′ = z + x ⋅ z′

173.

xy ′ − y = xy ′ =

x2 +y2

x2 +y2 +y

y′ = 1 +

y2 x

2

+

y x

y = z y ′ = z + xz ′ x

z + xz ′ = 1 + z 2 + z xz ′ = 1 + z 2 xdz = 1 +z 2 dx dz dx = x 1 +z 2

∫

dz 1 +z

2

=∫

∫ dx x

ln z + 1 + z 2 = ln x +ln C ln

y y2 + 1 + 2 = ln C ⋅ x x x

y y2 + 1 + 2 =C ⋅ x x x y+

y 2 +x 2 =C ⋅ x x

y+

x 2 + y 2 =C ⋅ x 2

175.

− 2 zdz dx −∫ =∫ 2 x 1− z

y + y ′ = x − dx dx y =  ∫ ⋅ ∫ ∫ ⋅ x dx +C     

y2 ln x + ln C + ln 1 − 2 = 0 x ln x ⋅ C ln  1=

2

x −y x2

2

x

y = −x ⋅

=0

x ⋅ C x2 − y2

= 0

2

y

C ⋅ x2 − y2 x 2

x = C ⋅ x − y2

y y y

176.

ln 1 − y

−1

3 C1 = x 1− y 3

− x3

y = 1 − C1 ⋅ 

y = 1 + C ⋅ − x

104

3

⋅ x dx +C

[

]

g ( x ) = x

f( x) = 1

− dx dx y =  ∫  C + ∫ ∫ ⋅ x dx   

(

3

y = − x C + ∫ 2 x dx

)

1   y = − x  C + 2 x  2  

pošto je y= 1 , za x=0 imamo 1=C +

C1 ⋅ − x = 1 − y

x

]

    2 x =t   2x  ∫  dx = 2dx = dt =  = −x ⋅ ∫ 2 x dx +C  dt dx =   2   1  1 2x t  ∫  dt =   2 2  1  = −x ⋅  2 x +C  2  1 C = x + x 2  x  C = + x 2 

y ′ + y = x

x3 +C 3

= ln x

[∫

177.

y ′ + 3 x 2 y = 3x 2 dy = 3 x 2 (1 − y ) dx dy = 3 x 2 dx ∫ 1− y − ln 1 − y = 3

g ( x ) = x

f ( x ) =1

− ln 1 − z 2 = ln x + C

1 2

tj.

C=

1 2

traženo partikularno rješenje je x − − x y= 2

178. y′ + 2 y = x 2 + 2x g( x) = x 2 + 2x

f( x) = 2

− f dx f dx y =  ∫ ( x ) ∫ ( x ) ⋅ g ( x ) dx + C   

(

)

− 2 dx  − 2 dx y = ∫ ⋅ ∫ ∫ ⋅ x 2 + x dx + C   

[

(

)

y = −2 x ⋅ ∫ 2 x ⋅ x 2 + x dx + C

]

 x 2 2 x x2 x 32 x  y = −2 x ⋅  + − + C  2 4  2 

(

)

 y = 2 x x 2 + 2 x dx = x 2 2 x dx + 2 x2 x dx =  ∫ ∫ ∫   2 2 x   x = u  dx = v ′   2 2x   1 2x  =  ∫ x  dx =   2 xdx = u ′ v =      2    2x     ′ x = u  dx = v 2 2x  = x  − 2 x xdx =  =  1 2x  ∫    2 dx = du v =    2     2 2x 2 2x x 2 2 x x2 x 2 x 1 2x  2x = = x  − x  + 1 2 x = − − + x −    2 2 4 2 2 4 2   x = u 2 x dx = v ′    =  ∫ 2 x2 x dx = 2 ∫ x2 x dx =   dx = du v = 1 2 x      2     2 x    = 2 ⋅  x − 1 2 x dx  = x ⋅ 2 x − 1 2 x  ∫  2    2 2      

179. y ′ cos x = y sin x + cos 2 x y ′ cos x − y sin x = cos 2 x / : cos x sin x y′ − y = cos x cos x y ′ − tgx ⋅ y = cos x f ( x ) = −tgx

g ( x ) = cos x

− f dx f dx y =  ∫ ( x ) ⋅  ∫ ∫ ( x ) ⋅ g ( x ) dx + C    tgxdx  − tgxdx y = ∫ ⋅ ∫  ∫ ⋅ cos xdx + C   

cos x = t  sin x dt   ∫ tgxdx = ∫ cos x dx = − sin xdx = dt  = −∫ t = − ln t = − ln cos sin xdx = − dt 

105

− ln cos x

y=

[

⋅ ∫

⋅ cos xdx + C

[

−1

ln cos x

y=

ln cos x

⋅ ∫ cos x ⋅ cos xdx + C

[

]

]

]

1 ⋅ ∫ cos 2 xdx + C cos x 1  1 + cos 2 x  y= ⋅ ∫ dx + C  cos x  2  y=

1 1 1  ⋅  x + sin 2 x + C  cos x  2 4  x sin 2 x y= + +C 2 cos x 4 cos x y=

180. xdy − 2 ydx = x 3 ln xdx / : dx dy x − 2 y = x 3 ln x dx x ⋅ y ′ − 2 y = x 3 ln x / : x 2 y ′ = y = x 2 ln x x 2 f( x) = − , g ( x ) = x 2 ln x x − f dx f dx u = ln x y =  ∫ ( x ) ⋅  ∫ ∫ ( x ) ⋅ g ( x ) dx + C    ln xdx = 1 ∫ u ′ = dx 2 2   − ∫ dx x ∫ dx y =  x ⋅  ∫  x ⋅ x 2 ln xdx + C   

(

)

[

y = 2 ln x ⋅ ∫ −2 ln x ⋅ x 2 ln xdx + C 2

[

−2

y = ln x ⋅ ∫ ln x ⋅ x 2 ln xdx + C

[ ⋅ [ ∫ ln xdx + C ]

y = x 2 ⋅ ∫ x − 2 ⋅ x 2 ln xdx + C y = x2

]

]

dx = v ′ v=x

= x ln x − ∫ dx = x ln x − x

]

y = x 2 ⋅ [ x ln x − x + C ]

181.

Konstanta za drugi izvod r 2

y ′′ − a y =  2

bx

r 2 − a2 = 0 polinom od r r1 = − a r2 = a

( p( r ) )

y 0 = C1 ⋅ − ax + C 2 ⋅ ax

η=

bx bx = 2 p( b ) b − a 2 − ax

y = y 0 + η = C1 ⋅  106

bx + C2 ⋅  + 2 b − a2 ax

182.

183.

y ′′ − 2 y ′ + y = 2 x

y ′′ − 3 y ′ + 2 y = 2 x x 2 + x

2

p : r − 2r + 1 = 0 r1, 2 = 1 1x

(

)

2

r − 3r + 2 = 0 1x

y 0 = C1 ⋅  + x ⋅ C 2 ⋅  b=2 p( b ) = p( 2) p( 2) = 2 2 − 2 ⋅ 2 + 1 = 1 2 x 2 x η= = = 2 x p( 2) 1

3± 9− 4⋅2 r1 = 1 r2 = 2 b = 2 2 Kada je b jednako sa jednim r1 ili r2 tada je η = x ⋅ ( a 0 + a1 x ) ⋅ 2 x r1, 2 =

y 0 = C1 ⋅ r1x + C 2 ⋅ r2 x y 0 = C1 ⋅ x + C 2 ⋅ 2 x

y = y 0 + η = C1 ⋅ x + x ⋅ C 2 ⋅ x + 2 x

184.

y +1 y′ = x dy y + 1 = dx x dy dx = y +1 x

ln y + 1 = ln x + ln C y +1 = C ⋅ x y = Cx − 1

y 1 = x x 1 1 y′ − ⋅ y = x x 1 1 f( x) = − g( x) = x x dx  1  − ∫ dx 1 ∫ y =  x ⋅  ∫  x ⋅ dx + C  x   −1 1   y = ln x ⋅  ∫ ln x ⋅ dx + C  x    1 1  y = x ⋅  ∫ ⋅ ⋅ dx + C   x x  y′ −

[

y = x ⋅ ∫ x − 2 dx + C  1  y = x ⋅ − + C   x  y = Cx − 1

107

]

Sadržaj stranica Funkcija – definicija ................................................................................ 3 Granične vrijednosti – ( limesi) .............................................................. 10 Granične vrijednosti funkcije ................................................................. 29 Diferencijalni račun ................................................................................. 35 Izvod .......................................................................................................... 36 Izvod složenih funkcija ............................................................................ 37 Funkcije .................................................................................................... 40 Integrali .................................................................................................... 75 Parcijalna integracija .............................................................................. 85 Određeni integrali ................................................................................... 95 Newton – Leibnitz-ova formula ............................................................. 95 Diferencijalne jednačine ......................................................................... 100 Sadržaj ..................................................................................................... 108

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