00:01
Here we're looking at a problem where we have to evaluate the integral of the ln of x squared.
00:08
So let's write that out, ln of x squared.
00:13
Now this can be written out like that, or it can be written out like this.
00:20
So these mean the same thing.
00:22
There's different ways of writing it.
00:24
So i'm going to write it out like this way in the right, because that's how i am used to doing it.
00:30
So let's continue on with this problem.
00:33
If we look at this, we're taking the integral of the ln of x squared.
00:40
So our first step would be picking our u, which we could make just ln squared, well, ln of x squared.
00:50
So let's write that out.
00:51
We write u equals ln of x squared, and then we can make our dv just one.
01:01
So it equals 1, and then our b would end up equaling x, and now we find our du.
01:09
And our du is going to be equal to, let's take the, you can look at this to make this simpler, so we're taking the derivative of it.
01:18
So we take our 2, lnx, and ln of x, and then we'd have to take the derivative of the inside, and the derivative of ln of x is just one of x so we get this whole thing is over x so now we have the parts that we need to get moving with this solution we take our uv which we have ln of x squared and just x let's write that out we have x ln of x squared minus our b which is x and then d u which is x and then d u which is 2 l.
02:04
Of x over x.
02:07
So our x is end up canceling out...