_-substitution: defining _ | AP Calculus AB | Khan Academy

What we're going to do in this video is give ourselves some practice in the first step of u substitution, which is often the most difficult for those who are first learning it. That's recognizing when u substitution is appropriate and then defining an appropriate u.

So let's just start with an example here. Let's say we want to take the indefinite integral of (2x + 1) times the square root of (x^2 + x , dx). Does u substitution apply here? And if it does, how would you define that u? Pause the video and try to think about that.

Well, we just have to remind ourselves that u substitution is really trying to undo the chain rule. If we remind ourselves what the chain rule tells us, it says look, if we have a composite function, let's say (F(G(x))), and we take the derivative of that with respect to (x), that is going to be equal to the derivative of the outside function with respect to the inside function, so (f'(G(x))) times the derivative of the inside function.

So u substitution is all about, well, do we see a pattern like that inside the integral? Do we see a potential inside function (G(x)) where I see its derivative being multiplied? Well, we see that over here. If I look at (x^2 + x), if I make that the (u), what's the derivative of that?

Well, the derivative of (x^2 + x) is (2x + 1), so we should make that substitution. If we say (u) is equal to (x^2 + x), then we could say (\frac{du}{dx}), the derivative of (u) with respect to (x), is equal to (2x + 1).

If we treat our differentials like variables or numbers, we can multiply both sides by (dx), which is a little bit of hand-wavy mathematics, but it's appropriate here. So we could say (2x + 1) times (dx).

Now what's really interesting is here we have our (u) right over there. Notice we have our (2x + 1 , dx). In fact, it's not conventional to see an integral rewritten the way I'm about to write it, but I will.

I could rewrite this integral—you should really view this as the product of three things. Oftentimes, people just view the (dx) as somehow part of the integral operator, but you could rearrange it. This would actually be legitimate; you could say the integral of the square root of (x^2 + x) times (2x + 1 , dx).

And if you wanted to be really clear, you could even put all of those things in parentheses or something like that. So here, this is our (U), and this right over here is our (DU).

We could rewrite this as being equal to the integral of the square root of (U) because (x^2 + x) is (U), times (DU), which is much easier to evaluate. If you are still confused, you might recognize it if I rewrite this as (u^{\frac{1}{2}}) because now we could just use the reverse power rule to evaluate this.

Then, we would have to undo the substitution. Once we figure out what this antiderivative is, we would then reverse substitute the (X) expression back in for the (U).