Integrace Prima
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Prˇ´ıma´ metoda integrace Robert Marˇ´ık a Lenka Bara´kova´ 16. srpna 2005
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Lenka Bara´kova´, 2005 ×
Obsah Z Z
Z
Z Z Z Z Z
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√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx x x+3 dx . . . . . . . . . . . . . . x2 tg x dx . . . . . . . . . . . . . . . 4x dx . . . . . . . . . . . . . x2 − 5 x+2 dx . . . . . . . . . . x2 + 4x + 5 1 dx . . . . . . . . . . 2 x + 2x + 3 1 dx . . . . . . . . . . . . ( x + 6)3
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3
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20
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26
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34
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38
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42
f ( ax + b) dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
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..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
• Integra´l ze soucˇtu je soucˇet integra´lu˚. • Integra´l na´sobku funkce je na´sobek integra´lu. //
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Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Vytkneme konstantu prˇed integra´l. //
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Vytkneme konstantu a prˇepı´sˇeme do mocninne´ funkce. //
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Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Vytkneme konstantu a prˇepı´sˇeme do mocninne´ funkce. //
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Vytkneme konstantu −1.
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Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x −2 x2 x5/4 +6 − (− cos x ) + e x + c +3 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Z //
x n dx = /
.
..
x n +1 +c n+1 c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x2 x5/4 x −2 − (− cos x ) + e x + c +3 +6 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Z //
x n dx = /
.
..
x n +1 +c n+1 c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x2 x −2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 1 12 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Z //
x n dx = /
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..
x n +1 +c n+1 c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 1 12 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Z //
sin x dx = − cos x + c /
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 1 12 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Z //
e x dx = e x + c /
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 1 12 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Upravı´me. //
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Upravı´me. //
/
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Upravı´me. //
/
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Upravı´me. //
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Upravı´me. //
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+3 dx. x2 Z
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x 3 + 2 dx x2 x Z Z 1 1 dx + 3 dx = x x2 x −1 = ln | x | + 3 · +c −1 3 = ln | x | − + c x
x+3 dx = x2
Z
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+3 dx. x2 Z
x 3 + 2 dx x2 x Z Z 1 1 = dx + 3 dx x x2 x −1 = ln | x | + 3 · +c −1 3 = ln | x | − + c x
x+3 dx = x2
Z
Pro integrova´nı´ je vhodneˇjsˇ´ı soucˇet, proto zlomek rozdeˇlı´me na dva. //
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Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+3 dx. x2 Z
x 3 + 2 dx x2 x Z Z 1 1 = dx + 3 dx x x2 x −1 = ln | x | + 3 · +c −1 3 = ln | x | − + c x
x+3 dx = x2
Z
Kazˇdy´ scˇ´ıtanec integrujeme zvla´sˇt’, konstanty vytkneme prˇed integra´l. //
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+3 dx. x2 Z
Z
1 dx = ln | x | + c x
Z //
x 3 + 2 dx x2 x Z Z 1 1 dx + 3 dx = x x2 x −1 = ln | x | + 3 · +c −1 3 = ln | x | − + c x
x+3 dx = x2
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Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+3 dx. x2 Z
Z //
x n dx = /
.
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x 3 + 2 dx x2 x Z Z 1 1 dx + 3 dx = x x2 x −1 = ln | x | + 3 · +c −1 3 = ln | x | − + c x
x+3 dx = x2
Z
x n +1 +c n+1 c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+3 dx. x2 Z
x 3 + 2 dx x2 x Z Z 1 1 dx + 3 dx = x x2 x −1 = ln | x | + 3 · +c −1 3 = ln | x | − + c x
x+3 dx = x2
Z
Nakonec vy´raz upravı´me. //
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
tg x dx. Z
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sin x dx cos x Z − sin x =− dx cos x Z (cos x )0 =− dx cos x = − ln | cos x | + c
tg x dx =
Z
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
tg x dx. Z
sin x dx cos x Z − sin x =− dx cos x Z (cos x )0 =− dx cos x = − ln | cos x | + c
tg x dx =
Z
V prˇ´ıpadeˇ, zˇe je v integra´lu funkce tangens vzˇdy jej rozepisujeme pomocı´ sinus a cosinus. //
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
tg x dx. Z
sin x dx cos x Z − sin x =− dx cos x Z (cos x )0 =− dx cos x = − ln | cos x | + c
tg x dx =
Z
• Platı´ (cos x )0 = − sin x. Cˇitatel se tedy lisˇ´ı od derivace jmenovatele jenom konstantı´m na´sobkem. • Vyna´sobı´me a vydeˇlı´me integra´l tı´mto na´sobkem. //
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Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
tg x dx. Z
sin x dx cos x Z − sin x =− dx cos x Z (cos x )0 =− dx cos x = − ln | cos x | + c
tg x dx =
Z
Forma´lneˇ pouzˇijeme vztah (cos x ) 0 = − sin x, abychom videˇli vzorec Z 0 f (x) dx = ln | f ( x )| + c. f (x) //
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Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
tg x dx. Z
Z
f 0 (x) dx = ln | f ( x )| + c f (x)
Z //
sin x dx cos x Z − sin x =− dx cos x Z (cos x )0 =− dx cos x = − ln | cos x | + c
tg x dx =
/
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
4x dx. x2 − 5 Z
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4x 2x dx = 2 dx x2 − 5 x2 − 5 = 2 ln | x2 − 5| + c Z
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
4x dx. x2 − 5 Z
4x 2x dx = 2 dx x2 − 5 x2 − 5 = 2 ln | x2 − 5| + c Z
V prˇ´ıpadeˇ jednoduche´ ryze lomene´ raciona´lnı´ funkce je vhodne´ pouzˇ´ıt Z 0 f (x) vzorec dx = ln | f ( x )| + c. Funkce f ( x ) = x 2 − 5, proto v f (x) cˇitateli potrˇebujeme f 0 ( x ) = 2x. Vytkneme prˇed integra´l 2. //
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Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
4x dx. x2 − 5 Z
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4x 2x dx = 2 dx x2 − 5 x2 − 5 = 2 ln | x2 − 5| + c Z
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+2 dx. x2 + 4x + 5 Z
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1 2x + 4 x+2 dx = dx 2 x2 + 4x + 5 x2 + 4x + 5 Z 1 ( x2 + 4x + 5)0 dx = 2 x2 + 4x + 5 1 = ln( x2 + 4x + 5) + c 2 Z
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+2 dx. x2 + 4x + 5 Z
1 2x + 4 x+2 dx = dx 2 x2 + 4x + 5 x2 + 4x + 5 Z 1 ( x2 + 4x + 5)0 dx = 2 x2 + 4x + 5 1 = ln( x2 + 4x + 5) + c 2 Z
• Platı´ ( x2 + 4x + 5)0 = 2x + 4. Cˇitatel se tedy lisˇ´ı od derivace jmenovatele jenom konstantı´m na´sobkem. • Vyna´sobı´me a vydeˇlı´me integra´l tı´mto na´sobkem. //
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Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+2 dx. x2 + 4x + 5 Z
1 2x + 4 x+2 dx = dx 2 x2 + 4x + 5 x2 + 4x + 5 Z 1 ( x2 + 4x + 5)0 dx = 2 x2 + 4x + 5 1 = ln( x2 + 4x + 5) + c 2
Prˇepı´sˇeme do tvaru //
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Z
Z
f 0 (x) dx. f (x) c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+2 dx. x2 + 4x + 5 Z
Z
f 0 (x) dx = ln | f ( x )| + c f (x)
Z //
1 2x + 4 x+2 dx = dx 2 x2 + 4x + 5 x2 + 4x + 5 Z 1 ( x2 + 4x + 5)0 dx = 2 x2 + 4x + 5 1 = ln( x2 + 4x + 5) + c 2
/
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
1 dx. x2 + 2x + 3 Z
//
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1 1 1 dx = dx 2 x2 + 2x + 3 ( x + 1)2 + 2 x+1 1 = √ arctg √ + c 2 2 Z
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
1 dx. x2 + 2x + 3 Z
1 1 1 dx = dx 2 x2 + 2x + 3 ( x + 1)2 + 2 x+1 1 = √ arctg √ + c 2 2 Z
Tentokra´t prˇedchozı´ postup nelze pouzˇ´ıt. V cˇitateli je konstanta. Z 1 1 x Pouzˇijeme proto vzorec dx = arctg 2 2 A A x +A //
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Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
1 dx. x2 + 2x + 3 Z
1 1 1 dx = dx 2 x2 + 2x + 3 ( x + 1)2 + 2 x+1 1 = √ arctg √ + c 2 2 Z
Jmenovatel prˇepı´sˇeme do tvaru ( x + neˇco)2 + zbyla´ konstanta. Tomuto triku rˇ´ıka´me doplneˇnı´ na cˇtverec: x2 + ax + b = //
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x+
a 2
2
−
a2 +b 4 c
Lenka Bara´kova´, 2005 ×
Najdeˇte
1 dx. x2 + 2x + 3
Z
Z
1 1 1 dx = dx 2 x2 + 2x + 3 ( x + 1)2 + 2 x+1 1 = √ arctg √ + c 2 2
Nynı´ pouzˇijeme vzorec f (x) =
x2
1 , tedy +2
Z //
/
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Z
Z
f ( ax + b) dx =
1 F ( ax + b) + c pro a
1 x+1 f ( x + 1) dx = F ( x + 1) + c = √ arctg √ + c. 2 2 c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
1 dx. ( x + 6)3 I=
Z
1 dx ( x + 6)3
=
Z
( x + 6)−3 dx
( x + 6) −2 −2 1 =− +C 2( x + 6)2
=
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/
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
1 dx. ( x + 6)3 I=
Z
1 dx ( x + 6)3
=
Z
( x + 6)−3 dx
( x + 6) −2 −2 1 =− +C 2( x + 6)2
=
Jedna´ se o mocninnou funkci. //
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
1 dx. ( x + 6)3 I=
Z
1 dx ( x + 6)3
=
Z
( x + 6)−3 dx
( x + 6) −2 −2 1 =− +C 2( x + 6)2 =
Z
•
f ( ax + b) dx =
1 F ( ax + b), kde F je integra´l z f . a
• V nasˇem prˇ´ıpadeˇ je f ( x ) = x −3 , F ( x ) = //
/
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x −2 a a = 1. −2 c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
1 dx. ( x + 6)3 I=
Z
1 dx ( x + 6)3
=
Z
( x + 6)−3 dx
( x + 6) −2 −2 1 =− +C 2( x + 6)2 =
Upravı´me. //
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. 1 1 dx = ln |2x + 5| + C 2x + 5 2 Z Z 1 dx = (2 − 1 · x )−5 dx (2 − x )5
Z
(2 − x ) −4 1 · −4 −1 1 = +C 4(2 − x )4
=
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Z
e− x dx = −e− x + C
Z
e3x dx =
1 3x e +C 3
c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. 1 1 dx = ln |2x + 5| + C 2x + 5 2 Z Z 1 dx = (2 − 1 · x )−5 dx (2 − x )5
Z
(2 − x ) −4 1 · −4 −1 1 = +C 4(2 − x )4
=
//
/
Z
e− x dx = −e− x + C
Z
e3x dx =
•
Z
1 dx = ln | x | x
•
Z
f ( ax + b) dx =
.
..
1 3x e +C 3
1 F ( ax + b), v nasˇem prˇ´ıpadeˇ a = 2. a c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. 1 1 dx = ln |2x + 5| + C 2x + 5 2 Z Z 1 dx = (2 − 1 · x )−5 dx (2 − x )5 Z
(2 − x ) −4 1 · −4 −1 1 = +C 4(2 − x )4
=
Z
e− x dx = −e− x + C
Z
e3x dx =
1 3x e +C 3
Prˇepı´sˇeme na mocninnou funkci. //
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. 1 1 dx = ln |2x + 5| + C 2x + 5 2 Z Z 1 dx = (2 − 1 · x )−5 dx (2 − x )5 Z
(2 − x ) −4 1 · −4 −1 1 = +C 4(2 − x )4 =
Z
//
/
•
Z
•
Z .
e− x dx = −e− x + C
1 3x = e3x + C 1 e ndx 3 +1 x dx = x n+1 n
Z
f ( ax + b) dx = ..
1 F ( ax + b), v nasˇem prˇ´ıpadeˇ a = −1. a c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. 1 1 dx = ln |2x + 5| + C 2x + 5 2 Z Z 1 dx = (2 − 1 · x )−5 dx (2 − x )5 Z
(2 − x ) −4 1 · −4 −1 1 = +C 4(2 − x )4
=
//
/
.
..
Z
e− x dx = −e− x + C
Z
e3x dx =
1 3x e +C 3
c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. 1 1 dx = ln |2x + 5| + C 2x + 5 2 Z Z 1 dx = (2 − 1 · x )−5 dx (2 − x )5 Z
(2 − x ) −4 1 · −4 −1 1 = +C 4(2 − x )4
=
//
/
Z
e− x dx = −e− x + C
Z
e3x dx =
•
Z
e dx = e
•
Z
f ( ax + b) dx =
.
x
..
x
1 3x e +C 3
1 F ( ax + b), v nasˇem prˇ´ıpadeˇ a = −1. a c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. 1 1 dx = ln |2x + 5| + C 2x + 5 2 Z • e x dx = e x Z 1 dx = (2 − 1 · x )−5 dx (2 − x )5 Z 1 • f ( ax + b) dx = F ( ax + b), v (2nas −ˇxem )−4prˇ´ıpade 1 ˇ a = 3. a = · −4 −1 1 = +C 4(2 − x )4 Z
Z
//
/
.
..
Z
e− x dx = −e− x + C
Z
e3x dx =
1 3x e +C 3
c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. Z
(e x + e− x )2 dx =
Z
(e2x + 2 + e−2x ) dx
1 1 2x e + 2x − e−2x + C 2 2 Z Z 1 1 1 sin(2x ) dx = · · (− cos 2x ) + C sin x cos x dx = 2 2 2 Z Z i 1 1h 1 1 − cos(2x ) dx = x − sin(2x ) + C sin2 x dx = 2 2 2
=
//
/
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. Z
(e x + e− x )2 dx =
Z
(e2x + 2 + e−2x ) dx
1 1 2x e + 2x − e−2x + C 2 2 Z Z 1 1 1 sin(2x ) dx = · · (− cos 2x ) + C sin x cos x dx = 2 2 2 Z Z i 1 1h 1 1 − cos(2x ) dx = x − sin(2x ) + C sin2 x dx = 2 2 2
=
Upravı´me podle vzorce ( a + b)2 :
(e x + e− x )2 = e2x + 2e x e− x + e−2x = e2x + 2 + e−2x //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. Z
(e x + e− x )2 dx =
Z
(e2x + 2 + e−2x ) dx
1 2x 1 e + 2x − e−2x + C 2 2 Z Z 1 1 1 sin(2x ) dx = · · (− cos 2x ) + C sin x cos x dx = 2 2 2 Z Z i 1 1h 1 2 = 1 − cos ( 2x ) dx = x − sin ( 2x ) +C sin x dx Integrujeme podle vzorcu 2 ˚ 2 2
=
Z Z Z //
/
.
..
f ( ax + b) dx =
e x dx = e x , 1 dx = x ,
1 F ( ax + b) , kde a
Z
f ( x ) dx = F ( x ). c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. Z
(e x + e− x )2 dx =
Z
(e2x + 2 + e−2x ) dx
1 2x 1 e + 2x − e−2x + C 2 2 Z Z 1 1 1 sin(2x ) dx = · · (− cos 2x ) + C sin x cos x dx = 2 2 2 Z Z i 1 1h 1 1 − cos(2x ) dx = x − sin(2x ) + C sin2 x dx = 2 2 2
=
Pouzˇijeme vzorec //
/
.
..
sin(2x ) = 2 sin x cos x c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. Z
(e x + e− x )2 dx =
Z
(e2x + 2 + e−2x ) dx
1 2x 1 e + 2x − e−2x + C 2 2 Z Z 1 1 1 sin(2x ) dx = · · (− cos 2x ) + C sin x cos x dx = 2 2 2 Z Z i 1 1h 1 1 − cos(2x ) dx = x − sin(2x ) + C sin2 x dx = 2 2 2
=
Integrujeme podle vzorcu˚ Z
a Z //
/
.
..
f ( ax + b) dx =
sin x dx = − cos x 1 F ( ax + b) , kde a
Z
f ( x ) dx = F ( x ). c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. Z
(e x + e− x )2 dx =
Z
(e2x + 2 + e−2x ) dx
1 2x 1 e + 2x − e−2x + C 2 2 Z Z 1 1 1 sin(2x ) dx = · · (− cos 2x ) + C sin x cos x dx = 2 2 2 Z Z i 1 1h 1 sin2 x dx = 1 − cos(2x ) dx = x − sin(2x ) + C 2 2 2
=
Vzorec sin2 x = //
/
.
..
1 − cos(2x ) 2 c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. Z
(e x + e− x )2 dx =
Z
(e2x + 2 + e−2x ) dx
1 2x 1 e + 2x − e−2x + C 2 2 Z Z 1 1 1 sin(2x ) dx = · · (− cos 2x ) + C sin x cos x dx = 2 2 2 Z Z i 1 1h 1 sin2 x dx = 1 − cos(2x ) dx = x − sin(2x ) + C 2 2 2
=
Z Z //
/
.
..
cos x dx = sin x
f ( ax + b) =
1 F ( ax + b) a c
Lenka Bara´kova´, 2005 ×
KONEC
//
/
.
..
c
Lenka Bara´kova´, 2005 ×
//
/
.
..
c
Lenka Bara´kova´, 2005 ×
Obsah Z Z
Z
Z Z Z Z Z
//
/
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx x x+3 dx . . . . . . . . . . . . . . x2 tg x dx . . . . . . . . . . . . . . . 4x dx . . . . . . . . . . . . . x2 − 5 x+2 dx . . . . . . . . . . x2 + 4x + 5 1 dx . . . . . . . . . . 2 x + 2x + 3 1 dx . . . . . . . . . . . . ( x + 6)3
. . . . . . . . . . . . . . . . . .
3
. . . . . . . . . . . . . . . . . .
20
. . . . . . . . . . . . . . . . . .
26
. . . . . . . . . . . . . . . . . .
31
. . . . . . . . . . . . . . . . . .
34
. . . . . . . . . . . . . . . . . .
38
. . . . . . . . . . . . . . . . . .
42
f ( ax + b) dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
//
/
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
• Integra´l ze soucˇtu je soucˇet integra´lu˚. • Integra´l na´sobku funkce je na´sobek integra´lu. //
/
.
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Vytkneme konstantu prˇed integra´l. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Vytkneme konstantu a prˇepı´sˇeme do mocninne´ funkce. //
/
.
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Vytkneme konstantu a prˇepı´sˇeme do mocninne´ funkce. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Vytkneme konstantu −1.
//
/
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
//
/
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x −2 x2 x5/4 +6 − (− cos x ) + e x + c +3 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Z //
x n dx = /
.
..
x n +1 +c n+1 c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x2 x5/4 x −2 − (− cos x ) + e x + c +3 +6 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Z //
x n dx = /
.
..
x n +1 +c n+1 c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x2 x −2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 1 12 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Z //
x n dx = /
.
..
x n +1 +c n+1 c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 1 12 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Z //
sin x dx = − cos x + c /
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 1 12 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Z //
e x dx = e x + c /
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 1 12 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Upravı´me. //
/
.
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Upravı´me. //
/
.
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Upravı´me. //
/
.
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Upravı´me. //
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.
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte Z
Z
√ 6 (2x + 3 4 + 3 − sin x + e x ) dx. x
√ 6 (2x + 3 4 x + 3 − sin x + e x ) dx xZ Z =2
x dx + 3
1
x 4 dx + 6
Z
x −3 dx −
Z
sin x dx +
Z
e x dx
x5/4 x −2 x2 +3 +6 − (− cos x ) + e x + c 2 5/4 −2 12 1 = x2 + x5/4 − 3 2 + cos x + e x + c 5 x
=2
Upravı´me. //
/
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+3 dx. x2 Z
//
/
.
..
x 3 + 2 dx x2 x Z Z 1 1 dx + 3 dx = x x2 x −1 = ln | x | + 3 · +c −1 3 = ln | x | − + c x
x+3 dx = x2
Z
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+3 dx. x2 Z
x 3 + 2 dx x2 x Z Z 1 1 = dx + 3 dx x x2 x −1 = ln | x | + 3 · +c −1 3 = ln | x | − + c x
x+3 dx = x2
Z
Pro integrova´nı´ je vhodneˇjsˇ´ı soucˇet, proto zlomek rozdeˇlı´me na dva. //
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+3 dx. x2 Z
x 3 + 2 dx x2 x Z Z 1 1 = dx + 3 dx x x2 x −1 = ln | x | + 3 · +c −1 3 = ln | x | − + c x
x+3 dx = x2
Z
Kazˇdy´ scˇ´ıtanec integrujeme zvla´sˇt’, konstanty vytkneme prˇed integra´l. //
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+3 dx. x2 Z
Z
1 dx = ln | x | + c x
Z //
x 3 + 2 dx x2 x Z Z 1 1 dx + 3 dx = x x2 x −1 = ln | x | + 3 · +c −1 3 = ln | x | − + c x
x+3 dx = x2
/
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+3 dx. x2 Z
Z //
x n dx = /
.
..
x 3 + 2 dx x2 x Z Z 1 1 dx + 3 dx = x x2 x −1 = ln | x | + 3 · +c −1 3 = ln | x | − + c x
x+3 dx = x2
Z
x n +1 +c n+1 c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+3 dx. x2 Z
x 3 + 2 dx x2 x Z Z 1 1 dx + 3 dx = x x2 x −1 = ln | x | + 3 · +c −1 3 = ln | x | − + c x
x+3 dx = x2
Z
Nakonec vy´raz upravı´me. //
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Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
tg x dx. Z
//
/
.
..
sin x dx cos x Z − sin x =− dx cos x Z (cos x )0 =− dx cos x = − ln | cos x | + c
tg x dx =
Z
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
tg x dx. Z
sin x dx cos x Z − sin x =− dx cos x Z (cos x )0 =− dx cos x = − ln | cos x | + c
tg x dx =
Z
V prˇ´ıpadeˇ, zˇe je v integra´lu funkce tangens vzˇdy jej rozepisujeme pomocı´ sinus a cosinus. //
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Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
tg x dx. Z
sin x dx cos x Z − sin x =− dx cos x Z (cos x )0 =− dx cos x = − ln | cos x | + c
tg x dx =
Z
• Platı´ (cos x )0 = − sin x. Cˇitatel se tedy lisˇ´ı od derivace jmenovatele jenom konstantı´m na´sobkem. • Vyna´sobı´me a vydeˇlı´me integra´l tı´mto na´sobkem. //
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Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
tg x dx. Z
sin x dx cos x Z − sin x =− dx cos x Z (cos x )0 =− dx cos x = − ln | cos x | + c
tg x dx =
Z
Forma´lneˇ pouzˇijeme vztah (cos x ) 0 = − sin x, abychom videˇli vzorec Z 0 f (x) dx = ln | f ( x )| + c. f (x) //
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Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
tg x dx. Z
Z
f 0 (x) dx = ln | f ( x )| + c f (x)
Z //
sin x dx cos x Z − sin x =− dx cos x Z (cos x )0 =− dx cos x = − ln | cos x | + c
tg x dx =
/
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
4x dx. x2 − 5 Z
//
/
.
..
4x 2x dx = 2 dx x2 − 5 x2 − 5 = 2 ln | x2 − 5| + c Z
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
4x dx. x2 − 5 Z
4x 2x dx = 2 dx x2 − 5 x2 − 5 = 2 ln | x2 − 5| + c Z
V prˇ´ıpadeˇ jednoduche´ ryze lomene´ raciona´lnı´ funkce je vhodne´ pouzˇ´ıt Z 0 f (x) vzorec dx = ln | f ( x )| + c. Funkce f ( x ) = x 2 − 5, proto v f (x) cˇitateli potrˇebujeme f 0 ( x ) = 2x. Vytkneme prˇed integra´l 2. //
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
4x dx. x2 − 5 Z
//
/
.
..
4x 2x dx = 2 dx x2 − 5 x2 − 5 = 2 ln | x2 − 5| + c Z
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+2 dx. x2 + 4x + 5 Z
//
/
.
..
1 2x + 4 x+2 dx = dx 2 x2 + 4x + 5 x2 + 4x + 5 Z 1 ( x2 + 4x + 5)0 dx = 2 x2 + 4x + 5 1 = ln( x2 + 4x + 5) + c 2 Z
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+2 dx. x2 + 4x + 5 Z
1 2x + 4 x+2 dx = dx 2 x2 + 4x + 5 x2 + 4x + 5 Z 1 ( x2 + 4x + 5)0 dx = 2 x2 + 4x + 5 1 = ln( x2 + 4x + 5) + c 2 Z
• Platı´ ( x2 + 4x + 5)0 = 2x + 4. Cˇitatel se tedy lisˇ´ı od derivace jmenovatele jenom konstantı´m na´sobkem. • Vyna´sobı´me a vydeˇlı´me integra´l tı´mto na´sobkem. //
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+2 dx. x2 + 4x + 5 Z
1 2x + 4 x+2 dx = dx 2 x2 + 4x + 5 x2 + 4x + 5 Z 1 ( x2 + 4x + 5)0 dx = 2 x2 + 4x + 5 1 = ln( x2 + 4x + 5) + c 2
Prˇepı´sˇeme do tvaru //
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.
..
Z
Z
f 0 (x) dx. f (x) c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
x+2 dx. x2 + 4x + 5 Z
Z
f 0 (x) dx = ln | f ( x )| + c f (x)
Z //
1 2x + 4 x+2 dx = dx 2 x2 + 4x + 5 x2 + 4x + 5 Z 1 ( x2 + 4x + 5)0 dx = 2 x2 + 4x + 5 1 = ln( x2 + 4x + 5) + c 2
/
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
1 dx. x2 + 2x + 3 Z
//
/
.
..
1 1 1 dx = dx 2 x2 + 2x + 3 ( x + 1)2 + 2 x+1 1 = √ arctg √ + c 2 2 Z
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
1 dx. x2 + 2x + 3 Z
1 1 1 dx = dx 2 x2 + 2x + 3 ( x + 1)2 + 2 x+1 1 = √ arctg √ + c 2 2 Z
Tentokra´t prˇedchozı´ postup nelze pouzˇ´ıt. V cˇitateli je konstanta. Z 1 1 x Pouzˇijeme proto vzorec dx = arctg 2 2 A A x +A //
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.
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
1 dx. x2 + 2x + 3 Z
1 1 1 dx = dx 2 x2 + 2x + 3 ( x + 1)2 + 2 x+1 1 = √ arctg √ + c 2 2 Z
Jmenovatel prˇepı´sˇeme do tvaru ( x + neˇco)2 + zbyla´ konstanta. Tomuto triku rˇ´ıka´me doplneˇnı´ na cˇtverec: x2 + ax + b = //
/
.
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x+
a 2
2
−
a2 +b 4 c
Lenka Bara´kova´, 2005 ×
Najdeˇte
1 dx. x2 + 2x + 3
Z
Z
1 1 1 dx = dx 2 x2 + 2x + 3 ( x + 1)2 + 2 x+1 1 = √ arctg √ + c 2 2
Nynı´ pouzˇijeme vzorec f (x) =
x2
1 , tedy +2
Z //
/
.
..
Z
Z
f ( ax + b) dx =
1 F ( ax + b) + c pro a
1 x+1 f ( x + 1) dx = F ( x + 1) + c = √ arctg √ + c. 2 2 c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
1 dx. ( x + 6)3 I=
Z
1 dx ( x + 6)3
=
Z
( x + 6)−3 dx
( x + 6) −2 −2 1 =− +C 2( x + 6)2
=
//
/
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
1 dx. ( x + 6)3 I=
Z
1 dx ( x + 6)3
=
Z
( x + 6)−3 dx
( x + 6) −2 −2 1 =− +C 2( x + 6)2
=
Jedna´ se o mocninnou funkci. //
/
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..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
1 dx. ( x + 6)3 I=
Z
1 dx ( x + 6)3
=
Z
( x + 6)−3 dx
( x + 6) −2 −2 1 =− +C 2( x + 6)2 =
Z
•
f ( ax + b) dx =
1 F ( ax + b), kde F je integra´l z f . a
• V nasˇem prˇ´ıpadeˇ je f ( x ) = x −3 , F ( x ) = //
/
.
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x −2 a a = 1. −2 c
Lenka Bara´kova´, 2005 ×
Najdeˇte
Z
1 dx. ( x + 6)3 I=
Z
1 dx ( x + 6)3
=
Z
( x + 6)−3 dx
( x + 6) −2 −2 1 =− +C 2( x + 6)2 =
Upravı´me. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. 1 1 dx = ln |2x + 5| + C 2x + 5 2 Z Z 1 dx = (2 − 1 · x )−5 dx (2 − x )5
Z
(2 − x ) −4 1 · −4 −1 1 = +C 4(2 − x )4
=
//
/
.
..
Z
e− x dx = −e− x + C
Z
e3x dx =
1 3x e +C 3
c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. 1 1 dx = ln |2x + 5| + C 2x + 5 2 Z Z 1 dx = (2 − 1 · x )−5 dx (2 − x )5
Z
(2 − x ) −4 1 · −4 −1 1 = +C 4(2 − x )4
=
//
/
Z
e− x dx = −e− x + C
Z
e3x dx =
•
Z
1 dx = ln | x | x
•
Z
f ( ax + b) dx =
.
..
1 3x e +C 3
1 F ( ax + b), v nasˇem prˇ´ıpadeˇ a = 2. a c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. 1 1 dx = ln |2x + 5| + C 2x + 5 2 Z Z 1 dx = (2 − 1 · x )−5 dx (2 − x )5 Z
(2 − x ) −4 1 · −4 −1 1 = +C 4(2 − x )4
=
Z
e− x dx = −e− x + C
Z
e3x dx =
1 3x e +C 3
Prˇepı´sˇeme na mocninnou funkci. //
/
.
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c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. 1 1 dx = ln |2x + 5| + C 2x + 5 2 Z Z 1 dx = (2 − 1 · x )−5 dx (2 − x )5 Z
(2 − x ) −4 1 · −4 −1 1 = +C 4(2 − x )4 =
Z
//
/
•
Z
•
Z .
e− x dx = −e− x + C
1 3x = e3x + C 1 e ndx 3 +1 x dx = x n+1 n
Z
f ( ax + b) dx = ..
1 F ( ax + b), v nasˇem prˇ´ıpadeˇ a = −1. a c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. 1 1 dx = ln |2x + 5| + C 2x + 5 2 Z Z 1 dx = (2 − 1 · x )−5 dx (2 − x )5 Z
(2 − x ) −4 1 · −4 −1 1 = +C 4(2 − x )4
=
//
/
.
..
Z
e− x dx = −e− x + C
Z
e3x dx =
1 3x e +C 3
c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. 1 1 dx = ln |2x + 5| + C 2x + 5 2 Z Z 1 dx = (2 − 1 · x )−5 dx (2 − x )5 Z
(2 − x ) −4 1 · −4 −1 1 = +C 4(2 − x )4
=
//
/
Z
e− x dx = −e− x + C
Z
e3x dx =
•
Z
e dx = e
•
Z
f ( ax + b) dx =
.
x
..
x
1 3x e +C 3
1 F ( ax + b), v nasˇem prˇ´ıpadeˇ a = −1. a c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. 1 1 dx = ln |2x + 5| + C 2x + 5 2 Z • e x dx = e x Z 1 dx = (2 − 1 · x )−5 dx (2 − x )5 Z 1 • f ( ax + b) dx = F ( ax + b), v (2nas −ˇxem )−4prˇ´ıpade 1 ˇ a = 3. a = · −4 −1 1 = +C 4(2 − x )4 Z
Z
//
/
.
..
Z
e− x dx = −e− x + C
Z
e3x dx =
1 3x e +C 3
c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. Z
(e x + e− x )2 dx =
Z
(e2x + 2 + e−2x ) dx
1 1 2x e + 2x − e−2x + C 2 2 Z Z 1 1 1 sin(2x ) dx = · · (− cos 2x ) + C sin x cos x dx = 2 2 2 Z Z i 1 1h 1 1 − cos(2x ) dx = x − sin(2x ) + C sin2 x dx = 2 2 2
=
//
/
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. Z
(e x + e− x )2 dx =
Z
(e2x + 2 + e−2x ) dx
1 1 2x e + 2x − e−2x + C 2 2 Z Z 1 1 1 sin(2x ) dx = · · (− cos 2x ) + C sin x cos x dx = 2 2 2 Z Z i 1 1h 1 1 − cos(2x ) dx = x − sin(2x ) + C sin2 x dx = 2 2 2
=
Upravı´me podle vzorce ( a + b)2 :
(e x + e− x )2 = e2x + 2e x e− x + e−2x = e2x + 2 + e−2x //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. Z
(e x + e− x )2 dx =
Z
(e2x + 2 + e−2x ) dx
1 2x 1 e + 2x − e−2x + C 2 2 Z Z 1 1 1 sin(2x ) dx = · · (− cos 2x ) + C sin x cos x dx = 2 2 2 Z Z i 1 1h 1 2 = 1 − cos ( 2x ) dx = x − sin ( 2x ) +C sin x dx Integrujeme podle vzorcu 2 ˚ 2 2
=
Z Z Z //
/
.
..
f ( ax + b) dx =
e x dx = e x , 1 dx = x ,
1 F ( ax + b) , kde a
Z
f ( x ) dx = F ( x ). c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. Z
(e x + e− x )2 dx =
Z
(e2x + 2 + e−2x ) dx
1 2x 1 e + 2x − e−2x + C 2 2 Z Z 1 1 1 sin(2x ) dx = · · (− cos 2x ) + C sin x cos x dx = 2 2 2 Z Z i 1 1h 1 1 − cos(2x ) dx = x − sin(2x ) + C sin2 x dx = 2 2 2
=
Pouzˇijeme vzorec //
/
.
..
sin(2x ) = 2 sin x cos x c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. Z
(e x + e− x )2 dx =
Z
(e2x + 2 + e−2x ) dx
1 2x 1 e + 2x − e−2x + C 2 2 Z Z 1 1 1 sin(2x ) dx = · · (− cos 2x ) + C sin x cos x dx = 2 2 2 Z Z i 1 1h 1 1 − cos(2x ) dx = x − sin(2x ) + C sin2 x dx = 2 2 2
=
Integrujeme podle vzorcu˚ Z
a Z //
/
.
..
f ( ax + b) dx =
sin x dx = − cos x 1 F ( ax + b) , kde a
Z
f ( x ) dx = F ( x ). c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. Z
(e x + e− x )2 dx =
Z
(e2x + 2 + e−2x ) dx
1 2x 1 e + 2x − e−2x + C 2 2 Z Z 1 1 1 sin(2x ) dx = · · (− cos 2x ) + C sin x cos x dx = 2 2 2 Z Z i 1 1h 1 sin2 x dx = 1 − cos(2x ) dx = x − sin(2x ) + C 2 2 2
=
Vzorec sin2 x = //
/
.
..
1 − cos(2x ) 2 c
Lenka Bara´kova´, 2005 ×
Najdeˇte na´sledujı´cı´ integra´ly. Z
(e x + e− x )2 dx =
Z
(e2x + 2 + e−2x ) dx
1 2x 1 e + 2x − e−2x + C 2 2 Z Z 1 1 1 sin(2x ) dx = · · (− cos 2x ) + C sin x cos x dx = 2 2 2 Z Z i 1 1h 1 sin2 x dx = 1 − cos(2x ) dx = x − sin(2x ) + C 2 2 2
=
Z Z //
/
.
..
cos x dx = sin x
f ( ax + b) =
1 F ( ax + b) a c
Lenka Bara´kova´, 2005 ×
KONEC
//
/
.
..
c
Lenka Bara´kova´, 2005 ×
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