Chapter 7

Additional Integration Topics

Section 3

Integration by Parts

2

Learning Objectives for Section 7.3 Integration by Parts

The student will be able to integrate by parts.

3

An Unsolved Problem

In a previous section we said we would return later to the indefinite integral

ln xdxsince none of the integration techniques considered up to that time could be used to find an antiderivative for ln x. We now develop a very useful technique, called integration by parts that will enable us to find not only the preceding integral, but also many others.

4

Integration by Parts

Integration by parts is based on the product formula for derivatives:

We rearrange the above equation to get:

Integrating both sides yields

which reduces to

d

dxf (x) g (x) f (x) g (x) g (x) f (x)

f (x) g (x)

d

dxf (x) g (x) g (x) f (x)

f (x) g (x)

d

dxf (x) g (x) g (x) f (x)

f (x) g (x) f (x) g (x) g (x) f (x)

5

Integration by Parts Formula

If we let u = f (x) and v = g(x) in the preceding formula, the equation transforms into a more convenient form:

Integration by Parts Formula

udv uv vdu

6

Integration by Parts Formula

This method is useful when the integral on the left side is difficult and changing it into the integral on the right side makes it easier.

Remember, we can easily check our results by differentiating our answers to get the original integrals.

Let’s look at an example.

duvvudvu

7

Example

Our previous method of substitution does not work. Examining the left side of the integration by parts formula yields two possibilities.

Let’s try option 1.

dxex x

Option 2

dxxe x

u dv

Option 1

dxex x

u dv

8

Example(continued)

We have decided to use u = x and dv = e x dx.

This means du = dx and v = e x, yielding

This is easy to check by differentiating.

Ceex

dxeexdxex

xx

xxx

xxxxxx exeex eCex edx

d

Note: Option 2 does not work in this example.

9

Selecting u and dv

■ The product u dv must equal the original integrand.

■ It must be possible to integrate dv by one of our known methods.

■ The new integral should not be more involved than the

original integral

■ For integrals involving x p e ax , try u = x p and dv = e axdx

■ For integrals involving x p (ln x) q, try u = (ln x) q and dv = x p dx

duvvudvu

duv

dvu

10

Example 2

duvvudvu dxxx ln3 usingFind

11

Example 2

Let u = ln x and dv = x3 dx, yielding du = 1/x dx and v = x4/4

Check by differentiating:

duvvudvu dxxx ln3

Cx

xx

dxx

xx

xdxxx 16

ln4

1

4ln

4ln

44443

xxx

xxx

x

C x

x – x

dx

dln

16

4ln

1

416ln

43

33

444

usingFind

12

Solving the Unsolved Problem

Recall that our original problem was to find

We follow the suggestion

For integrals involving x p (ln x)q, try u = (ln x)q and dv = x p dx

Set p = 0 and q = 1 and choose u = ln x and dv = dx.Then du = 1/x dx and v = x, hence

1ln (ln )( ) ( )

ln

xdx x x x dxx

x x x C

ln xdx

13

Summary

We have developed integration by parts as a useful technique for integrating more complicated functions.

Integration by Parts Formula:

duvvudvu