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# 17.3 Integration by Parts

17.17   Theorem (Integration by parts.) Let be functions that are continuous on an interval and differentiable on . If has an antiderivative on , then also has an antiderivative on and
 (17.18)

We call formula (17.18) the formula for integration by parts.

Proof:     This theorem is just a restatement of the product rule for differentiation. If and are differentiable on then the product rule says that

so that

on . If is an antiderivative for on , then is continuous on , and

on . We have shown that is an antiderivative for on .

17.19   Example. We will calculate . We begin by searching for an antiderivative for . Let

Then by the formula for integration by parts
 (17.20)

Hence

Suppose I had tried to find in the following way: Let

Then by the formula for integration by parts
 (17.21)

In this case the antiderivative looks more complicated than the one I started out with. When you use integration by parts, it is not always clear what you should take for and for . If you find things are starting to look more complicated rather than less complicated, you might try another choice for and .

Integration by parts is used to evaluate antiderivatives of the forms
, , and when is a positive integer. These can be reduced to antiderivatives of the forms ,
, and , so by applying the process times we get the power of down to , which gives us antiderivatives we can easily find.

17.22   Example. We will calculate

Let

Then
 (17.23)

We will now use integration by parts to find an antiderivative for . Let

Then
 (17.24)

From equations (17.23) and (17.24) we see that

Thus

and

17.25   Example. We will calculate . Let

Then

17.26   Theorem (Antiderivative of inverse functions.) Let and be intervals in , and let be a continuous function such that is defined and non-zero for all in the interior of . Suppose that is the inverse function for , and that is an antiderivative for . Then an antiderivative for on is given by
 (17.27)

Proof: Let . Then , and

 (17.28)

Now and for all in , so

and it follows from (17.28) that

Remark: It follows from the proof of the previous theorem that if you know an antiderivative for a function , then you can find an antiderivative for the inverse function by integration by parts. This is what you should remember about the theorem. The formula (17.27) is not very memorable.

17.29   Example. In the previous theorem, if we take

then we get

This agrees with the result obtained in example 17.25.

17.30   Exercise. What is wrong with the following argument? Let

Then

If we subtract from both sides we obtain

17.31   Exercise. A Calculate the following antiderivatives:
a)
b)
(Integrate by parts more than once.)
c)
d)
e)
f)
, where . Have you considered the case where ?
g)
.

Next: 17.4 Integration by Substitution Up: 17. Antidifferentiation Techniques Previous: 17.2 Basic Formulas   Index
Ray Mayer 2007-09-07