Continuing on the path of reversing derivative rules in order to make them useful for integration, we reverse the product rule.
If
where
and
are functions of
, then

Rearranging,

Therefore,

Therefore,

or

This is the integration by-parts formula. It is very useful in many integrals involving products of functions and others.
For instance, to treat

we choose
and
. With these choices, we have
and
, and we have

Note that the choice of
and
was critical. Had we chosen the reverse, so that
and
, the result would have been

The resulting integral is no easier to work with than the original; we might say that this application of integration by parts took us in the wrong direction.
So the choice is important. One general guideline to help us make that choice is, if possible, to choose
as the factor of the integrated, which becomes simpler when we differentiate it. In the last example, we see that
does not become simpler when we differentiate it:
is no more straightforward than
.
An important feature of the integration-by-parts method is that we often need to apply it more than once. For instance, to integrate

we start by choosing
and
to get

Note that we still have an integral to take care of, and we do this by applying integration by parts again, with
and
, which gives us

So, two applications of integration by parts were necessary, owing to the power of
in the integrand.
Note that any power of x does become simpler when we differentiate it, so when we see an integral of the form

one of our first thoughts ought to be to consider using integration by parts with
. Of course, for it to work, we need to be able to write down an antiderivative for
.
Use integration by parts to evaluate the integral

Solution: If we let
and
, then we have
and
. Using our rule for integration by parts gives

We do not seem to have made much progress.
But if we integrate by parts again with
and
and hence
and
, we obtain

We may solve this identity to find the anti-derivative of
and obtain

The rule is essentially the same for definite integrals as long as we keep the endpoints.
Integration by parts for definite integrals
Suppose f and g are differentiable and their derivatives are continuous. Then

.
This can also be expressed in Leibniz notation.

Examples Set 1: Integration by Parts
Evaluate the following using integration by parts.
Solutions