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We can think of **integration by parts** as a way to undo the product rule. While integration by substitution lets us find antiderivatives of functions that came from the chain rule, integration by parts lets us find antiderivatives of functions that came from the product rule.

Integration by parts requires learning and applying the integration-by-parts formula. Here's the formula, written in both Leibniz and Lagrange notation:

It's time to be skeptical. Why should

be true? Just because we wrote down an equation doesn't mean it's valid.

- What should be the derivative of with respect to
*x*?

- What should be the derivative of with respect to
*x*?

- What's the derivative of
*uv*?

Answer.

- is the family of all functions whose derivative is
*uv'*. If we take the derivative of any function in that family, we get*uv'*. So it makes sense for the derivative of to also be*uv'*. In symbols,

- is the collection of all functions whose derivative is
*vu'*. If we take the derivative of any one of those functions, we get*vu'*, so the derivative of should also be*vu'*. In symbols,

- This is a straightforward application of the product rule. The derivative of
*uv*is*uv'*+*u'v*.

Let's return to the question of why the equation

should be valid. Each side of this equation describes a family of functions. If each side of the equation describes the same family of functions, then the equation is valid.

describes the family of functions with derivative *uv*'.

The expression

describes the family of functions with derivative

Since *u'v* and *vu'* are the same, when we simplify we get

*u'v* + *uv'* – *vu'* = *u'v*.

This means the expression

also describes the family of functions with derivative *u'v*.

Since the expressions

and

describe the same family of functions, the equation

is valid.

Hopefully, you're now convinced that

To use this formula we have to

- pick
*u*and*v*',

- figure out what
*u '*and*v*are, and

- stick everything in the formula and simplify.

If we're using the notation

then we pick *u* and *dv*, figure out *du* and *v*, then apply the formula.

### Some Tricks

### There's Always a Factor of 1

We can use integration by parts to find the integral of something that doesn't look like a product. This is because whatever the integrand is, we can think of it as the product of itself and 1. Then we can choose

*v'*= 1 and apply the integration-by-parts formula.For example, since

ln

*x*= (ln*x*)(1),we know

If we chose

*u*= 1 then*u'*would be zero, which doesn't seem like a good idea. So take*u*= ln*x**v'*= 1### Factoring

Sometimes we need to rearrange the integrand in order to see what

*u*and*v*' should be. Exponents can be deceiving.### Sample Problem

For example, look at the integral

This looks like a product, so we want to use integration by parts. However, choosing

*u*=*x*^{5}or

*u*=*e*^{x3}won't work very well (try it yourself if you don't believe us; we're not going to demonstrate). But what if we rewrite the integrand by factoring

*x*^{5}?Now we can see it's reasonable to choose

*v'*=*x*^{2}*e*^{x3},since we can use substitution to figure out the antiderivative

*v*. This leaves*u*=*x*^{3}.### Integrating by Parts Twice

We already did some exercises where you had to integrate by parts twice: once to start off, then again to find the new integral. There are some problems where, if you integrate by parts twice, the original integral shows up again. Sometimes this will be as helpful as the equation

0 = 0

(that is, not helpful at all). However, sometimes when the original integral shows up again, you'll get an equation that you can rearrange to solve for the original integral.

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