# Indefinite Integrals

### Topics

## 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.