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.
Exercise 1
Apply the formula for integration by parts with u = ln x and v' = 1 to find
Exercise 2
Use integration by parts to find .
Exercise 3
Find
by applying the formula for integration by parts with u = x3 and v' = x2ex3.
Exercise 4
Find
Exercise 5
Use integration by parts to find
taking u = sin x for the first integration.
Exercise 6
Integrate.
Exercise 7
Integrate.
Exercise 8
Integrate.
Exercise 9
Integrate.
Exercise 10
Integrate.
Exercise 11
Integrate.
Exercise 12
Integrate.