00:01
We want to evaluate this definite integral here.
00:05
Now, you might notice that to the power of 2 here, we have natural log of x, and the derivative of natural log of x is 1 over x, and we have a 1 over x in this, so it may be that we had some chain rule going on.
00:21
So to integrate this, maybe we let u equal to natural log of x.
00:27
And that means du is going to be 1 over x dx.
00:33
So we can go ahead and fill all that in now.
00:37
So it would just be 2 to the u, du, and then our new bounds are going to be, well, when x is equal to 1, u is equal to the natural log of 1, which is 0...