 In this screencast, we're going to integrate the function e to the 2x with respect to x. The first thing to notice here is that this function is composite. It's easy to miss this, but notice the 2x in the exponent. That, because it's more than just a single variable, is enough to make this composite. So that means we'll need to use integration by substitution or u substitution in order to integrate this. So the goal with integration by substitution is to find an inner function and then also find its derivative multiplied on the outside. So in this case, there's really only one choice. The inner function has to be u, our name for it, is 2x. If you think of this as being the backwards chain rule, we've just identified the inside part of the composite function. Now, we also need the derivative of u with respect to x, which is called du dx, and that derivative is just 2. However, we usually rewrite this in differential notation, which looks like we're just bringing the dx over to the other side. Now, our goal here is to match every part of the integrand, the e to the 2x, with a u or part of the du. And you might notice we don't have any 2 sitting on the outside of the integrand. However, we can do a little bit of algebra and just divide by 2 to get 1 half du is equal to dx. And now we have everything we want. The u right here matches the 2x, so those two pink parts match. We also have the dx that we have to match, matching the dx over here. So now we're ready to integrate. So I'm going to start writing this out. I'm going to begin with what I'm given, the integral of e to the 2x dx, and then I'm going to perform the substitution replacing everything so that I never have u's and x's together in the same place. So I'm going to replace everything at once. So I'll start with e to the u because u is the same as 2x. And then for the dx, I'll substitute 1 half du. Alright, this is good. I've got all u's. Now I know that by the constant multiple rule for integrals, I can pull this constant multiple, the 1 half, out front. And now what I have left, the integral of e to the u du, is an elementary antiderivative. It's something that I can calculate directly. And I know that the general antiderivative of e to the u is e to the u, plus any constant, so I write the plus c. Alright, we're almost done, but we started with x's, so we need to end with x's. So I'm going to replace everything. I'm going to unsubstitute. So that's 1 half e, and u is 2x, plus a constant, which stays there. And this right here is my general antiderivative. Now the nice thing about integrating is you can always double check your work by differentiating. So let's double check this. I'm going to double check that the antiderivative I just found really is an antiderivative. I'll do that by taking the derivative with respect to x. So I have to use the chain rule here. That's a good sign that u substitution was the right choice. And that involves bringing down a 2, and the constant c turns into a 0. And when I simplify that 1 half and 2, I'm left with just e to the 2x, which is what I started with. And so that's a sign that I've done the right thing.