00:02
So let's try out this integral, starting off with a u substitution.
00:08
Let's try u equals 2x.
00:11
So du is going to be 2dx.
00:18
Sorry, let's not make u equal to 2x.
00:19
Let's make u equal to x squared.
00:21
We're trying to get rid of that ugly x squared.
00:23
So du is going to be 2x, dx.
00:28
So then this integral is going to become, so we know dx can be replaced with du over 2x, so du over 2x times 1 over, we have this u plus 1 term times that we still have this x.
00:52
So here's another x squared, so this is going to be the integral of du, we can pull out the half of u over u plus 1, which is perfect for partial fractions...