[MUSIC] Generally speaking, 
anti-differentiating things with radicals 
is difficult. 
For example, let's try to find an 
anti-derivative of, x over the cube root 
of x plus 1 dx right? 
Let's try to anti-differentiate this. 
Can I make a substitution to improve the 
integrand? 
So what substitution do I want to make? 
Well, I really have some choices, I mean, 
one thing I could do is I could make ux 
plus 1 and in that case, du is just dx. 
And if u is x plus 1, then x is u minus 
1, and I can then rewrite this 
anti-differentiation problem. 
instead this becomes the anti-derivative 
of that x, but u minus 1 still cube root, 
but instead of x plus 1, it's now of u 
and the dx becomes du. 
And I, I could actually do this, but I'm 
not going to head down that path. 
Instead, I want to show you a trick by 
which you can get rid of the radical all 
together. 
So instead let's set u, just to grab the 
whole denominator. 
So u will be the cube root of x plus 1 
and in that case, u cubed is x plus 1. 
Or, in other words, x is u cubed minus 1. 
And that tells me what dx is. 
In that case, dx is 3u squared du. 
Now I could make this substitution. 
This transforms this integration problem 
to, the anti-derivative of x now becomes 
u-cubed minus 1. 
The whole denominator just becomes u. 
And then the dx here is now a 3u squared 
du, so I'll write 3u squared du. 
And that's just a polynomial. 
So I'll simplify this a bit, this is the 
anti-derivative of, what is this here, u 
cubed over u times 3u squared, that's 3u 
to the 4th minus 1 over u times 3u 
squared that's minus 3u, du. 
And this, this is just a polynomial, so I 
can anti-differentiate this very quickly. 
This anti-derivative is 3 5ths u to the 
5th, minus 3 halves u squared plus C. 
Now, we'll just replace u with what it 
equals in terms of x. 
So u, remember, is the cube root of x 
plus 1, so putting that in here, I get 3 
5ths, instead of u it's x plus 1 to the 1 
3rd to the 5th. 
So to the 5 3rds power, minus 3 halves x 
plus 1 to the 1 3rd squared, which is to 
the 2 3rds power plus C. 
To make it easy to talk about this 
particular technique, it has a name. 
The rationalizing substitution. 
The original problem had a cube root in 
it, but I made a substitution to get rid 
of the radical, a rationalizing. 
A challenge when doing u substitutions is 
that there really are multiple paths that 
you can take to the correct answer. 
It's not even apparent how successful a 
particular substitution might be until 
after you've started applying it, right? 
You have to take that first bite before 
you realize whether or not you're eating 
food.