4.6

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Integration Techniques: Integration by Parts

OBJECTIVE• Evaluate integrals using the formula for integration by parts.

• Solve applied problems involving integration by parts.

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THEOREM 7

The Integration-by-Parts Formula

u dv uv v du

4.6 Integration Techniques: Integration by Parts

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Tips on Using Integration by Parts:1. If you have had no success using substitution, try

integration by parts.2. Use integration by parts when an integral is of the

form

Match it with an integral of the form

by choosing a function to be u = f (x), where f (x) can be differentiated, and the remaining factor to be dv = g (x) dx, where g (x) can be integrated.

f (x)g(x) dx .

u dv

4.6 Integration Techniques: Integration by Parts

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3. Find du by differentiating and v by integrating.4. If the resulting integral is more complicated than the

original, make some other choice for u = f (x) and dv = g (x) dx.

5. To check your solution, differentiate.

4.6 Integration Techniques: Integration by Parts

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Example 1: Evaluate: ln x dx.

4.6 Integration Techniques: Integration by Parts

1

and .du dx v xx

Then,

Let ln and .u x dv dx

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4.6 Integration Techniques: Integration by Parts

Example 1 (concluded): Then, the Integration-by-Parts Formula gives

lnx x x C

lnx x dx

ln x dx

1(ln )x x x dx

x

u dv uv v du

1ln and .u x and du dx v x

x

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4.6 Integration Techniques: Integration by Parts

Quick Check 1

Evaluate: 3 .xxe dx3Let and .xu x dv e dx

31 and .

3xdu dx v e Then,

Then, the integration by parts formula gives

3 3 31

3 3x x xx

xe e e dx

3 31

3 9x xx

e e C

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Example 2: Evaluate:

Let’s examine several choices for u and dv.

Attempt 1: This will not work because we do not know how to integrate

x ln x dx.

4.6 Integration Techniques: Integration by Parts

ln .dv x dx

Let and ln .u x dv x dx

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Example 2 (continued): Attempt 2:

Using the Integration-by-Parts Formula, we have

4.6 Integration Techniques: Integration by Parts

Let ln and .u x dv x dx

21

Then and .2x

du dx vx

( ln )x x dx

2 2 1ln

2 2

x xx dx

x

2 1

ln2 2

xx xdx

2 2

ln2 4

x xx C

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Example 3: Evaluate:

Using the Integration by Parts Formula gives us

x 5x 1 dx.

4.6 Integration Techniques: Integration by Parts

3 22Then 1 and 5 1 .

15du dx v x

1 2Let and (5 1) .u x dv x dx

( 5 1)x x dx 3 2 3 22 2(5 1) (5 1)

15 15x x x dx

3 2 5 22 2 2(5 1) (5 1)

15 25 15x x x C

3 2 5 22 4(5 1) (5 1)

15 375x x x C

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4.6 Integration Techniques: Integration by Parts

Quick Check 2

Evaluate: 2 3 2 .x x dxLet 2 and 3 2 .u x dv x dx

Then, 3/222 and 3 2 .

9du dx v x

Using the integration by parts formula, we get:

3/2 3/24 22 3 2 3 2 3 2 2

9 9x x dx x x x dx

3/2 5/24 43 2 3 2

9 135x x x C

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Example 4: Evaluate:

Note that we already found the indefinite integral in Example 1. Now we evaluate it from 1 to 2.

ln x dx1

2

4.6 Integration Techniques: Integration by Parts

2ln2 1

2ln2 2 0 1

(2 ln2 2) (1 ln1 1)

2

1ln x dx

2

1lnx x x

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Example 5: Evaluate to find the area of the shaded region shown below.

x2e xdx0

7

4.6 Integration Techniques: Integration by Parts

Then 2 and .xdu x dx v e

2Let and .xu x dv e dx

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Example 5 (continued):Using the Integration-by-Parts Formula gives us

To evaluate the integral on the right, we can apply integration by parts again, as follows.

4.6 Integration Techniques: Integration by Parts

2( )xx e dx 2( ) (2 )x xx e e x dx

2 2x xx e xe dx

Then 2 and .xdu dx v e

Let 2 and .xu x dv e dx

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Example 5 (continued):Using the Integration-by-Parts Formula again gives us

Then we can substitute this solution into the formula on the last slide.

4.6 Integration Techniques: Integration by Parts

2 2x xxe e C

2 ( )xx e dx 2 ( ) (2 )x xx e e dx

2( 2 2)xe x x C

2 2 2x x xx e xe e C 2 xx e dx

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Example 5 (concluded):Then, we can evaluate the definite integral.

4.6 Integration Techniques: Integration by Parts

1.94

765 2e

7 2 0 2(7 2 7 2) (0 2 0 2)e e

7 72 2

00( 2 2)x xx e dx e x x C

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4.6 Integration Techniques: Integration by Parts

Quick Check 3

Evaluate:3

0.

1

xdx

x 1

Let and .1

u x dv dxx

Then, 1/2 and 2 1 .du dx v x

Using the integration by parts formula, we get:

1/2 1/22 1 2 1

1

xdx x x x dx

x

1/2 3/242 1 1

3x x x C

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4.6 Integration Techniques: Integration by Parts

Quick Check 3 Concluded

Then we can evaluate the definite integral.

3

3 1/2 3/2

00

42 1 1

31

xdx x x x

x

1/2 3/2 1/2 3/24 42 3 3 1 3 1 2 0 0 1 0 1

3 3

1/2 3/2 3/24 46 4 4 1

3 3

4 8 412

3 3

36 32 4

3 3 3

8 2 or 2

3 3

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4.6 Integration Techniques: Integration by Parts

Section Summary

• The Integration-by-Parts Formula is the reverse of the Product Rule for differentiation:

• The choices for u and dv should be such that the integral is simpler than the original integral. If this does not turn out to be the case, other choices should be made.

v du

.u dv uv v du