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[MUSIC]. 

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Sometimes it isn't clear that you can do 
integration by parts until after you 

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perform some substitution. 
For example, let's attack this integral, 

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the integral of e to the square root of 
x, dx. 

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We'll start by making a substitution that 
looks totally unjustified. 

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But in retrospect, will turn out to be a 
brilliant move. 

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Well, here's the substitution I'm 
proposing. 

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Let's set u equal to the square root of 
x. 

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How does that change the integral? 
So in this case, if u is the square root 

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of x, then du is 1 over 2 square roots of 
x, dx. 

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in other words, right? 
I could solve this and find that dx is 2u 

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du. 
Now, what does that end up doing to the 

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integral? 
Well, this integral then becomes the 

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integral of e to the u, and instead of 
dx, it's now 2u du. 

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now it's a polynomial times an 
exponential function. 

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We can use parts. 
But, if I'm going to use parts, I 

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probably wanted to call them u and v, 
right? 

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But, I'll just change the names. 
Instead of talking about u and dv, I'll 

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talk about v and dw. 
So, in this problem I'll have v be 2u and 

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dw will be e to the u, du. 
And now the derivative of v, right? 

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Gives me that dv is 2 du. 
And an anti-derivative here I can pick 

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one, I'll have w be e to the u. 
So in that case, what does parts tell me? 

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Well, parts then tells me that this 
integral is the same as 2u, e to the u 

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minus the integral of w, dv, which is e 
to the u times 2 du. 

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But now, this in integration problem that 
I can do. 

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I know an anti-derivative of e to the u. 
So, this is 2u, e to the u minus 2e to 

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the u. 
I'm going to go back now. 

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And remember that I set u equal to the 
square root of x. 

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So, this gives me 2 square root of x, e 
to the square root of x minus 2e to the 

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square root of x. 
If I notice there's a common factor of 2e 

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to the square root of x, I could collect 
those terms and write this as 2e to the 

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square root of x times the square root of 
x minus 1. 

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And that's an anti-derivative for e to 
the square root of e. 

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And if I want to be a little bit more 
careful, maybe I'll include a plus C at 

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the end. 
We did it. 

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Contrast that with, say this integration 
problem, the integral of e and the 

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negative of x squared dx. 
This is an integration problem that I 

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can't solve using the functions that I 
have at hand. 

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And that this problem looks really 
similar to the original problem here, e 

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to the square root of x, right? 
It's e to some power of x. 

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So, you'd think that if we can solve this 
one, we should also be able to solve this 

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one. 
Seemingly similar problems, right? 

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Can have totally different outcomes.