Study Guide

Indefinite Integrals - Integration by Parts: Indefinite Integrals

Integration by Parts: Indefinite Integrals

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:

Why the Formula is True

It's time to be skeptical. Why should

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

Sample Problem

  • 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.

How to Use the Formula

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 = x5

    or

    u = ex3

    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 x5?

    Now we can see it's reasonable to choose

    v' = x2ex3,

    since we can use substitution to figure out the antiderivative v. This leaves

    u = x3.

    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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