[MUSIC]. 
Sometimes it isn't clear that you can do 
integration by parts until after you 
perform some substitution. 
For example, let's attack this integral, 
the integral of e to the square root of 
x, dx. 
We'll start by making a substitution that 
looks totally unjustified. 
But in retrospect, will turn out to be a 
brilliant move. 
Well, here's the substitution I'm 
proposing. 
Let's set u equal to the square root of 
x. 
How does that change the integral? 
So in this case, if u is the square root 
of x, then du is 1 over 2 square roots of 
x, dx. 
in other words, right? 
I could solve this and find that dx is 2u 
du. 
Now, what does that end up doing to the 
integral? 
Well, this integral then becomes the 
integral of e to the u, and instead of 
dx, it's now 2u du. 
now it's a polynomial times an 
exponential function. 
We can use parts. 
But, if I'm going to use parts, I 
probably wanted to call them u and v, 
right? 
But, I'll just change the names. 
Instead of talking about u and dv, I'll 
talk about v and dw. 
So, in this problem I'll have v be 2u and 
dw will be e to the u, du. 
And now the derivative of v, right? 
Gives me that dv is 2 du. 
And an anti-derivative here I can pick 
one, I'll have w be e to the u. 
So in that case, what does parts tell me? 
Well, parts then tells me that this 
integral is the same as 2u, e to the u 
minus the integral of w, dv, which is e 
to the u times 2 du. 
But now, this in integration problem that 
I can do. 
I know an anti-derivative of e to the u. 
So, this is 2u, e to the u minus 2e to 
the u. 
I'm going to go back now. 
And remember that I set u equal to the 
square root of x. 
So, this gives me 2 square root of x, e 
to the square root of x minus 2e to the 
square root of x. 
If I notice there's a common factor of 2e 
to the square root of x, I could collect 
those terms and write this as 2e to the 
square root of x times the square root of 
x minus 1. 
And that's an anti-derivative for e to 
the square root of e. 
And if I want to be a little bit more 
careful, maybe I'll include a plus C at 
the end. 
We did it. 
Contrast that with, say this integration 
problem, the integral of e and the 
negative of x squared dx. 
This is an integration problem that I 
can't solve using the functions that I 
have at hand. 
And that this problem looks really 
similar to the original problem here, e 
to the square root of x, right? 
It's e to some power of x. 
So, you'd think that if we can solve this 
one, we should also be able to solve this 
one. 
Seemingly similar problems, right? 
Can have totally different outcomes.