 All right, let's continue on with our lecture 16 about street-fighting integrals, right? In the previous video, we saw how algebraic techniques can be useful in helping us rewrite integrans in ways that are more preferable for integration. I mean, heck, that's what partial fraction decomposition is all about. So I wanna mention that a good rule of thumb is to try using U substitution before trying integration by parts. Integration by parts is a harder technique. If U substitution will work, use it first. In particular, I like to use integration by parts as a last resort, typically. The only exceptions is if I can see that integration by parts is an obvious technique to use here. Efforts first be employed to utilize any obvious substitution or manipulate the integrant to a form that's more appropriate for that U substitution using some type of algebraic or trigonometric identities. Certain function forms should be recognized, right? So for example, if the integrant is trigonometric, then a combination of trigonometric identities and U substitutions probably will be quite fruitful like we saw in section 7.2 of Stuart's textbook. If the function is rational and no convenient substitution is available, maybe decompose the rational function to partial fractions and then integrate it like we saw in section 7.4 of Stuart's textbook. If the integrand involves radicals, perhaps a trigonometric substitution would work or also this idea of a rationalizing substitution might be helpful here. This is what I wanna show you in this example. Let's try to integrate the square root of X plus four over X. This is a rational function that doesn't involve a square root here. And so our first inclination might be to do some type of trigonometric substitution, but if we're doing that, we'd have to take the square root of X to equal two tangent of that, two tangent theta, and that's gonna be really, really messy. That's not what we wanna do. So instead, we're gonna try what I call a rationalizing substitution or a ratio sub for short. And I'll explain the name in just a moment. So what you're gonna do for your rationalizing substitution, you're gonna take the square root and set it equal to U. Take the square root of X plus four as U. In which case, then take the derivative by the usual chain rule. You'll get DX over two times the square root of X plus four. And so the reason why this is useful is that square roots, and in fact, all radical functions, they're sort of a self-similarity between the original function and the square root. And that is its derivative, excuse me. You'll notice that both the square root and the original function has a square root, but then the derivative has that exact same square root inside of that. And we can make a substitution. This is U. And therefore, this expression looks like DU equals DX over two U. And with this perspective, when we cleared the denominators, we see that two UDU equals DX. And so we can make a substitution here. This is U. We can make the substitution that DX is equal to two UDU. And so with these substitutions in mind, you would end up with, again, the square root of X plus four is a U on top. The DX is a two UDU. What do you do with the X on the bottom? Well, these things right here are equations. You can manipulate them as equations. Square both sides, you get that U squared equals X plus four. So track four, you get X equals U squared minus four. And that's what you're gonna put in right here. U squared minus four. And so therefore, this thing looks like two U squared over U squared minus four DU. I'm gonna factor the two outside because it's just a constant multiple. I'm gonna squeeze it in right here. So you get this two integral of U squared over U squared minus four. And this is where the name rationalizing substitution gets its name. We've done a substitution. And I think there's two reasons one goes here. We did a substitution involving a square root right. The square root is this irrational expression. You'll notice that after the substitution, the square root's gone. So in some respect, we've like rationalized the fraction. So that's one way of thinking about it. Although I think the term actually derives from the idea that if you do this rationalize the substitution, you turn the function typically for which we can then use partial fraction decomposition to help us here. I mean, we have U squared over U squared minus four. Because this is, because we have a U squared on top of the bottom, I'm gonna do not long division, but short division to help me out here. Because we're gonna take U squared minus four over U squared minus four. And then since I subtracted four, I have to add four to go from there. U and like that. And so the first one, the first fraction will just simplify just to be a one. The other one, we're gonna have to do a little bit more with it. And so we end up with the integral. Oh, I switched to red there. We're gonna end up with the integral of two DU. And then with the second one, we're gonna have an eight, the integral of DU over U squared minus four. And there's two on, you could do a partial fraction decomposition, which is the approach I'm going to take right here. Cause after all, if you take eight over U squared minus four, you could break that up as A over U minus two plus B over U minus two. Clearing the denominators, you get eight equals A U minus two. Let's see, actually I wrote U minus two twice didn't die. We'll make the first one be negative, the second one be positive. So we get A times U plus two, and then we get B times U minus two. And then annihilate things, right? So when U equals negative two, you're gonna annihilate the A and you end up with A equals negative four B, therefore B equals negative two. When you plug in U equals two, that'll annihilate the B. And so you get eight equals four A, that is A equals two. So with that approach, you can end pretty quickly. Anti-derivative of two, of course is gonna be a two U. And this next one, we're going to get the integral of two over U minus two, DU, and then minus the integral of two over U plus two, DU. And then so by the usual trick we do here, you're gonna get two, the natural log of U minus two, minus two times the natural log of U plus two, plus a constant. You can actually combine those natural logs together. They're both a common coefficient of two. So you get two U plus two natural log of the absolute value of U minus two over U plus two, plus a constant, something like that. And then remove the U by remembering that it was just the square root of four, square root of X plus four, excuse me. So you get two times the square root of X plus four right here, and then you're gonna get two times the natural log of the absolute value of the square root of X plus four minus two over the square root of X plus four plus two, plus a constant. And so we can find the anti-derivative using partial fraction decomposition. It works out pretty nicely in this situation. If you wanted to though, back here, you could have also tried to do some type of like secant substitution. You could take U to equal two secant theta. Now it'd be okay. You could do that as well if you prefer the over the partial fraction decomposition. Because again, when it comes to street fighting, there might be more than one option. Do what works best for you, right? If you can do it, then you can do it. And that is what you're gonna use. If you can do it, you can use it. We'll say it that way.