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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.
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'.
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
describe the same family of functions, the equation
Hopefully, you're now convinced that
To use this formula we have to
If we're using the notation
then we pick u and dv, figure out du and v, then apply the formula.
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),
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
Sometimes we need to rearrange the integrand in order to see what u and v' should be. Exponents can be deceiving.
For example, look at the integral
This looks like a product, so we want to use integration by parts. However, choosing
u = x5
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.
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.