# Calculus/Integration techniques/Integration by Parts

Continuing on the path of reversing derivative rules in order to make them useful for integration, we reverse the product rule.

## Integration by parts[edit | edit source]

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 .

### Example[edit | edit source]

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

### With definite integral[edit | edit source]

The rule is essentially the same for definite integrals as long as we keep the endpoints.

Integration by parts for definite integralsSupposefandgare differentiable and their derivatives are continuous. Then

- .

This can also be expressed in Leibniz notation.

### More Examples[edit | edit source]

Examples Set 1: Integration by Parts

## Exercises[edit | edit source]

Evaluate the following using integration by parts.