[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.