so that

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

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

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.

Then

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

Then

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

Thus

and

Proof: Let . Then , and

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.

then we get

This agrees with the result obtained in example 17.25.

Then

If we subtract from both sides we obtain

- a)
- b)
- (Integrate by parts more than once.)
- c)
- d)
- e)
- f)
- , where . Have you considered the case where ?
- g)
- .