Chapter 8 Indefinite Integration Only

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Ch 8 Indefinite Integration

S6/KC

8.1 Primitive Functions and Indefinite Integrals (predict)


From differentiation,




Example 

d [F (x )] = f (x ) , where F(x) is the primitive / anti-derivative function of f(x) dx

1.

d 4 ( x ) = 4x3 dx

2.

d 4 x + 2 = 4x3 dx

(

)

d  4 1 3  x −  = 4x dx  2 There are many solution

3.

 



d [F (x ) + C ] = f (x ) -- (*) dx where [F ( x ) + C ] is the collection called infinite integrals of f (x) with respect to x. C is arbitrary constant And (*) becomes d [F ( x ) + C ] = f ( x )dx [from the definition of differential of the previous chapter.]




Notation



∫ f (x )dx = ∫ d [F (x ) + C ] = F (x ) + C ∫

where

is the integral sign, f (x) is the integrand,

dx is used to specify the integration w.r.t. x.




Integration is the reverse process of differentiation, and it is also called anti-differentiation.




Example 8.1.1

∫x

−4



Find



Analysis: predict that d x −3 = −3 x −4 dx

dx

then



−3

−4

dx

 1 −3  Since d  x  = x − 4 dx 3 −   Integrate both side w.r.t. x

∫x


( ) 1 d (x ) = x −3

−4

 1 −3  1 − 3 dx = ∫ d  x = x + C where C is arbitrary constant. −3  −3

After integration, the index of the polynomial will be increased by 1. But when index is equal to -1 of the integrand, the case is different.

P. 1/7

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Ch 8 Indefinite Integration

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Example 8.1.2 7

∫ x dx



Find



Analysis: d [ln ( x )] =



Since

1 dx x

d [7 ln ( x )] =

7 dx x

Integrated both sides w.r.t. x 7

∫ x dx = ∫ d [7 ln(x )] = 7 ln(x ) + C , where C is arbitrary constant.



Different format of integrand can result the same primitive function. Example 8.1.3



(a)

(EX 8.1, No. 30)

  ( x + 1)2 (x + 1) dx = (x + 1)dx d + C1  = 2 2   2

   x   x2 d  + x + C 2  = 2  + 1 dx = ( x + 1)dx   2  2 

(b) Expand

(x + 1)2

2 By Comparing, ∴ C2 =

+ C1 =

x2 1 + x + + C1 2 2

1 + C1 2

8.2 Formulae for Indefinite Integration


Basic formula: ∫ d [ f ( x )] = f ( x ) + C




Some standard formulae for indefinite integrals are listed as follows. Differentiation

( )

Integration 1  1 dx = ∫ d  x n  = x n + C , n ≠ −1 n  n

d n x = nx n −1 dx

∫x

d (kx ) = k d (x ) = k dx dx

∫ kdx = k ∫ dx = kx + C

Constant

d (ln x ) = 1 dx x

∫ x dx = ∫ dx[ln x ] = ln x + C

ln x

( )

n −1

1

1  1 dx = ∫ d  e kx  = e kx + C k  k

d kx e = ke kx dx

∫e

d [k ⋅ f (x )] = k d [ f (x )] dx dx

∫ k ⋅ f (x )dx = k ∫ f (x )dx

kz

Power’s Rule

ex Multiply by k

P. 2/7

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Ch 8 Indefinite Integration

d [u (x ) ± v(x )] = d [u (x )] ± d [v(x )] dx dx dx d dx

(uv ) = v

d dx

(u ) + u

d dx

(v )

d (log a x ) = 1 dx x ln a

Sum /

uv = ∫ vdu + ∫ udv

Product Rule

1

( )




∫ [u (x ) ± v(x )]dx = ∫ u(x )dx ± ∫ v(x )dx

ln x

∫ x ln a dx = ln a + C = log

d x a = a x ln a dx

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a

Difference

x+C

 ax  ax ∫ a dx = ∫ d  ln a  = ln a + C x

Example 8.2.1

 



1 1  + 6 dx x5 + e x − 7x3 + 3 2x  1  6 5 1 x  3 ∫  3 ⋅ x + 3 e − 7 x + 2 x + 6 dx

∫  3 ⋅

Evaluate

= 3∫ 6 x 5 dx +

6

1 x 1 1 e dx − 7 ∫ x 3 dx + ∫ dx + 6 ∫ dx ∫ 3 2 x

 5 +1   x6  1 x  x 3+1  1  + ln x + 6 x + C  = 3 e 7 + −  5 3 3 1 +  2   + 1  6  11

18 6 1 x 7 4 1 x + e − x + ln x + 6 x + C 11 3 4 2 [The integration can be separated into different parts.] =

8.3 Integration by Substitution
Sometimes it is much easier if the integration is done with respect to other pattern of variables. That pattern is called “substitution”.




Procedures using the substitution method u = u(x)






1.

Change the integrand in terms of u and du;

2.

Evaluate the new integral;

3.

replace u by x in the answer.

Example 8.3.1



Evaluate the followings (a) (b)

∫ (2 x + 1) dx 5

∫

e

x

x

dx

P. 3/7

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(a)

Ch 8 Indefinite Integration

∫ (2 x + 1) dx 5

u = 2x + 1

Substitute u and du into the original equation,

(b)

=

1 u 5+1 ⋅ +C 2 5 +1

=

1 (2 x + 1)6 + C 12

∫

e

[Substitute back the expressions of u and du]

x

x

u=e

dx

2du =

= 2u + C




x

x

du = e

= ∫ (2du )

= 2e

du = 2dx 1 du = dx 2

1  = ∫ u  du  2  5



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x

e

⋅

1 2 x

dx

x

x

dx

+C

The principle of the method of integration by substitution is chain rule.

d [F (u )] = d [F (u )] ⋅ du = f (u ) ⋅ g ′(x ) dx du dx d [F (u )] = f (u ) ⋅ g ′( x )dx F ( g ( x )) = ∫ d [F (u )] = ∫ f (u ) ⋅ g ′( x )dx [The substitution should be done COMPLETELY!! Don’t try to mix up with u and x.]




It is frequently used in the case of irrational functions in which the expression under the radical sign is




of the first degree, that is ax + b. The substitution can be u = ax + b or u = ax + b . Example 8.3.2



Evaluate



x2

x2

∫

x −1

dx

u = x −1 ⇒ x = u +1 du = dx

∫

x −1

=∫

(u + 1)2 du

dx

u

1 −1 1−   2− 1 = ∫  u 2 + 2u 2 + u 2 du   5

3

1

2 4 = u 2 + u 2 + 2u 2 + C 5 3 5 3 1 2 4 = ( x − 1) 2 + ( x − 1) 2 + 2( x − 1) 2 + C 5 3




Different integration methods only lead to the result different by constant.




Relationship in between should be studied. P. 4/7

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∫ (x

Evaluate

∫ (x

2 x 3 + 3x 4

2 x + 3x

+ 3x 2 + 7

3

4

+ 3x 2 + 7

)

4

dx = ∫

)

4

dx

1 1 1 ⋅ du = 3 + C 4 u 2 6u

u = x 4 + 3x 2 + 7 du = 4 x 3 + 6 x dx

(

(

(

)

1 du = 2 x 3 + 3 x dx 2

)

4

Example 8.3.4



Evaluate



∫2

3t

u = 2 3t du = (3)2 3t (ln 2 )dt

3t ∫ 2 dt

dt = ∫ u ⋅

1 1 du = u2 + C 3 ln 2 3 ln 2

( )

1 du = 2 3t dt 3 ln 2

2 3t +C 3 ln 2

=

Example 8.3.5 xdx  ∫ 2 x + 1 ln x 2 + 1

(

) (

(

)

=

(

)

u = ln x 2 + 1

2 xdx x2 +1 xdx 1 du = 2 2 x +1

du =

1 = ∫ du 2u 1 = ln u + C 2




)

1 +C 6 x + 3x 2 + 7

=




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Example 8.3.3






Ch 8 Indefinite Integration

)

1 ln ln x 2 + 1 + C 2

For this method, usually the complex part be substituted.

** Extra Notes on the Integration Method**


Expansion  Expand the polynomial before perform integration, as integration can be done part by part which is easier. Then simplified the index. 2



Example 1

 1  ∫  x + x  dx 2






 1  1 1 2  ∫  x + x  dx = ∫  x + 1 + x dx = 2 x + x + ln x + C

Rationalization P. 5/7

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Ch 8 Indefinite Integration

If the denominator of the function is in radical sign, simply it can be a great idea sometimes. Make use of the identity a 2 − b 2 = (a − b )(a + b ) i.e. x − y =



S6/KC



∫

1

∫

Example 2

x +1 + x

1 x +1 + x

(

x− y

)(

x+ y

)

dx

dx

3 3  x + 1 − x   x +1 − x 1 2 2 2   2  = ∫  dx = ∫ x + 1 − x dx = 3 ( x + 1) − 3 x + C  x + 1 + x  x + 1 − x 



Multiply the numerator and denominator by something.

 

dx

∫1+ e

Example 2.1

−x

dx 1 ex ex = ⋅ = dx ∫ 1 + e−x ∫ 1 + e−x e x ∫ e x + 1 dx 1 = ∫ du = ln u + C u

(

)

= ln e x + 1 + C




du = e x dx

[as ex + 1 must be greater than zero, so no absolute sign is needed.]

Integration by part (Out of the theme, but useful knowledge)  Sometimes differentiate the function, some of the relationship can be studied for better integration. 

∫ te

Example 3






u = ex +1

3t ∫ te dt =

3t

( )

dt

d te 3t = e 3t + 3te 3t

[ ( )

]

1 1 1 d te 3t − ∫ e 3t dt = te 3t − e 3t + C ∫ 3 3 9

te 3t =

[( )

1 d te 3t − e 3t 3

]

Partial Fraction  Refer to the notes of chapter 7  Example 4 x A B = + (a) Find A and B if 2 (x − 1) x − 1 (x − 1)2 x dx (b) Evaluate ∫ (x − 1)2 x A B A( x − 1) + B  (a) = + = 2 2 (x − 1) x − 1 (x − 1) (x − 1)2

x = A( x − 1) + B

--

(1)

Sub x = 1 into (1) 1=B Sub x = 0 into (1) 0 = -A + B A=1 P. 6/7

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(b)

Ch 8 Indefinite Integration

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

= ln x − 1 −




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1 +C x −1

Exponential function’s integration  Make sure both the numerator and the denominator got the expression ekx.

8.4 Applications of Indefinite Integrals


….EDITING….

P. 7/7