 Alright, let's talk about integration techniques. So far we've found antiderivatives in two different ways. Either we just recognize them, so the antiderivative of 1 over x, I think about that and say, well, what function has derivative 1 over x? Oh, that's a log of x, plus c, don't forget the constant of anti-differentiation. Or we can do guess and modify, so the antiderivative of 3x, well, that's an x raised to the first power, so I'm guessing there's going to be an x squared in the antiderivative, but if I differentiate x squared, I get 2x, not 3x, so I have to modify that by multiplying through by a constant, again, don't forget that constant of integration. Now if we want to do anything more complicated, we're going to have to develop some integration techniques. This is an important topic in second semester calculus, but there's at least one of the integration techniques that's worth introducing here. Just as a note and a preview of things to come, there's actually only two integration techniques. There's what's called a u-substitution, and then there's something called integration by part. Everything else comes down to algebra or trigonometry. So let's take a look at a u-substitution. So we want to find the antiderivative of e to the power of 5x dx. Now in a u-substitution, what we're going to do is we're going to sweep under the rug a complicated expression. What do we mean by a complicated expression? Well, it's helpful to remember one of the basic guidelines for the chain rule. The type of function that we're dealing with is the last thing that you do. And so the expression that we're going to sweep under the rug, at least initially, is something that's going to make it more clear what type of function we're dealing with. So let's take a look at it. This function here, e to the 5x, well it's actually an e to the something type of function. So the substitution that we might want to try is, well, let's make u equal whatever our something is. And again, we'll apply the kindergarten rule. We'll put everything back where we found it. So we had e to the 5x. And so what I want to do is I want to try u equals e to the 5x as our substitution. So remember, paper is cheap. So we'll go ahead and write everything down. We'll let u equal 5x. And I'm going to differentiate that. So my derivative is 5. And the reason that we want to do this is that our anti-derivative has this dx, this differential here. Now again, this is not a fraction. We can't really separate it. But notation exists and persists because it reminds us of something useful. And in this particular case, we can pretend that this is a fraction. And I can solve for dx in terms of everything else. Solving that equation, cross-multiplying if you want. du is equal to 5 times dx, which says that dx is 1 5th du. Now this is called a u substitution. So there's my u. And now we'll move on to the substitution part. I'm going to make that substitution if u equals 5x. Then this anti-derivative, u equals 5x. So I'll replace my 5x here with a u, my dx, 1 5th du. I'll do some really complicated algebra. And then I'll integrate. Now before we move on, here's an important check. Remember the differential controls the variable. So this differential, this du tells us that the only variable we're allowed to have is u. Let me check our expression here. The only variable here is u. And we're good. We're ready to find the anti-derivative. So that's a one-viv. That's a constant multiple. So I can factor that out in front of the anti-derivative. And, well, I do have to recognize some anti-derivatives. So let's think about this. I need a function whose derivative is e to the u. Well, that's the world's easiest anti-derivative. That's just going to be e to the u. And then don't forget our constant of integration. And again, apply the kindergarten rule. Put everything back where you found it. The important thing here is that our original anti-derivative is expressed in terms of x. We don't want to write our answer this way. This is expressed in terms of u. We don't want u in our final answer. We want x in our final answer. So I'll make that substitution back. u equals 5x. So I'll substitute back. And there's my final answer, 1 5th e to the 5x plus constant. Now, one last step. I'm not going to do this. But you should get into the habit of always doing this. When you find an anti-derivative, make sure that you check by differentiating and verifying that you actually get what you started with.