 Let's do a final example from lecture 15, which would be our last example about partial fraction decompositions. This one's gonna be a little bit of a doozy, so maybe pause the movie and take a bathroom break before we get started here. Because what's gonna happen this time is that we're gonna have an irreducible quadratic, x squared plus four. Now good news for us, this is already a completed square, it's a sum of two squares, so that saves us a lot of effort, we don't have to commit the square. But in this example, it's a repeated, it's a repeated irreducible quadratic. What does that do for the template here? So what's gonna happen is we're looking for things of the form ax plus b over x squared plus four plus cx plus d over x squared plus four squared, dx. This is the template that we have to look for. Because the quadratic is repeated, we need to have a partial fraction for the first power, and the second power. So we need to have two of those things. If we clear the denominators, we're gonna get x cubed plus x squared is equal to ax plus b times x squared plus four, and then we have cx plus d. Now if you want to, you can try to use annihilation to help us here, but it's gonna get really messy really quickly, it doesn't work super well. I would recommend just to foil these things out and try the systems of equations. So we're gonna get ax cubed plus four ax plus bx squared plus four b plus cx plus d. And so let's combine some like terms here. If we take an equation for x cubed, x cubed, you get a one on the left hand side. On the right hand side, we're just gonna get an a. So that's actually pretty nice, right? We already know what a is if we take the system of equation approach. And if you try x squared, on the left hand side, you're gonna get a one again that comes from this term. If you look for x squared on the right, you just find a b, in which case you get one equals b. So we already know what b is, that's pretty nice. If we try the x term on the right hand side, left hand side, there's nothing there, so you get a zero. On the right hand side, we're gonna get a four a and a c. So you get four a plus c. But I wanna admit that we already know what a is. a is equal to one. So this is actually, that is to say zero equals four plus c, which says that c equals negative four. That's not so bad at all. And then finally, if you look at the constant term, same basic idea, the left hand side doesn't have a constant, so that means it's zero. And for the right hand side, you have a four b and a d, so you actually get the exact same situation, zero equals four b plus d, and since b is one, this will tell us that d is equal to negative four. So even though we had a linear system with four equations, four unknowns, we're actually able to solve it very, very quickly and get one, one negative four and negative four. So what that tells us for our integral, so we had the first one had the denominator of x squared plus four, dx. And this is the one that was ax plus b, so it's just gonna give us x plus one. And then the next one, we had a coefficient, this is the one where you had a x squared plus four squared on the bottom. And this is the one that had negative four, so we're gonna get negative four x minus four dx like so. And so you can factor out the common, that common divisor negative four on the second one. So actually, I'm gonna write this as, get rid of that, get rid of that. We're gonna write this as negative four x plus one, like so. And so now we have the partial fractions here. And now the whole point of partial fractions is to get something that's more comfortable for more comfortable for the integration, how we can integrate this thing. So now we're gonna do a u substitution right here, because when we have these irreducible quadratics, we start with a u substitution. X squared plus four is our u. And so du is gonna equal two x, and that's it, two x dx. And that's gonna be the same u substitution we use here and in here, right? And so we want a two x in the numerator, so we can accomplish that by times e by two, divide by one half. This one already has a two, so we can kind of borrow it. We have a two right here and then a two right here. That's gonna give us one half, the integral of two x. If you distribute this two, of course, we don't want the two. So we have a two x dx over x squared plus four. And then we're gonna have a, well, you distribute the two, but there's also one half here. It's just gonna become the integral of dx over x squared plus four, like so. So the second one, we're gonna get negative two times the integral of two x dx over x squared plus four squared. And then the next one, we're gonna have negative four times dx over x squared plus four squared, like so. So we break these things up. So now the first two, sorry, I shouldn't say the first two, the first one and the third one are perfect for u substitution. So what we've seen already by the u substitution we did right here, this will look like, this first one would look like one over u. This one over here is gonna look like one over u squared. And so by u substitution, we're gonna get one half the natural log of x squared plus four, we get that one. Then for the second one by the power rule, that would become a negative one over u. So we're gonna get plus two, plus two over x squared plus four. So we can take care of those ones just fine. Now what do we do on this one right here? I told you in a previous video that we're gonna want to try to memorize anything of the form dx over x squared plus a squared. The anti-derivative looks like one over a arc tangent of x over a plus a constant. And so you just wanna use that one right here because it exactly has that template. And so we end up with, in this case, a is two, so we get one half arc tangent of x over two. And then the last thing we have to do with is this one, this four integral of dx over x squared plus four squared. Now this one we don't have it memorized. We'll actually just wanna have to do a trig sub right here. And so kind of going with that, whoops, going with this one, our trig sub, much like it's the trig sub we would have done for this thing if we went through all the details here, we would take x to equal two tangent theta, which then says that dx equals two secant squared theta d theta. And this also tells us that the square root of x squared plus four equals two secant theta, which means that x squared plus four equals four secant squared. All right, so plugging those into that expression, I'm just gonna deal with just this for a moment. Put it here. The dx, remember, becomes a two secant squared theta d theta. This, it's above x squared plus four, which is gonna be four secant squared theta. But then we're squaring that. So canceling some terms, there is a secant squared that's gonna cancel. And then you have a coefficient of one eighth when you're all done here. So we have one eighth the integral of just secant squared on the bottom. Now, since we just have a secant squared in the denominator, it's gonna be my recommendation. It's gonna be my recommendation you replace that with us just a cosine, right? Because secant on the bottom is a cosine on the top. So this would become one eighth the integral of cosine squared theta d theta. And when we have just a cosine squared, you wanna use a half angle identity. So remember that cosine squared is the same thing as one half, one plus cosine of two theta d theta, for which case we can quite easily anti-differentiate that. The one eighth and one half come together to make a one sixteenth. Anti-derivative one is theta and anti-derivative of cosine of two theta, that would be a one half sine of two theta. And then there's a constant right here. Now, when you have the sine of two theta, we wanna relate this in terms of theta, not two theta. So we're gonna write this as one sixteenth theta plus sine of two theta is the same thing as two sine theta, cosine theta. So you get sine theta, cosine theta plus a constant. The two cancels with the one half that was there. I never actually drew the triangle this time, but remember our triangle diagram would look something like the following. And so we have theta right here. Our original substitution remember was x was equal to two tangent theta, which is to say that tangent theta is x over two. So that's opposite over adjacent. So you get x over two. The other side will be the square root of x squared plus four. And so putting all of those together then, we're gonna get one sixteenth. For theta, we actually will get arctangent of x over two for sine and cosine, right? Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. So that ends up giving us a two x over x squared plus four. And so this is our anti-derivative of that very final integral. Let's take a look at what we had before. So there's an arctangent and there's a x squared over four part that shows up. You'll notice that we also have a arctangent of x over two and we also have a two over x squared plus four. So we make those substitutions, make sure you times this answer we just got by negative four. When you plug those in here, I have to kind of do it down here. When you plug those in, we had this one half the natural log of x squared plus four. We had this two over x squared plus four, remember? We had this one fourth arctangent of x over two. And then finally we then got this and I'm combining some light terms right there. And then maybe I shouldn't go too far ahead then. Let me back up a little bit. So remember we had a one half arctangent right there. So let me just copy that down, arctangent of x over two. And so from this red part, if we insert that into the problem, we can distribute these things through. We end up with a one sixteenth that I get my coefficient right there. And then I feel like something might be off. We get a one sixteenth, my spider sense is tingling, but we'll come back to look at that in a second, x over two. And then we had likewise this one sixteenth two x over x squared plus four plus a constant. Like so, I feel like this shouldn't be a 16. I feel like it should be much smaller than that. Let's come back to the red area. What was going on there? So we had a one eighth that's in front. Oh yes, I know what the issue is. I forget I forgot the one fourth and the negative one fourth that should be there as well. So these two guys, that's where my mistake is. These two guys right here should have a negative one fourth that sits in front of them. Negative fourth, of course, times by a sixteenth gives us a negative one fourth like there. And so as we then distribute these things through, we're gonna have a one, looking at just the arc tangents, we have a one half arc tangent. We have a negative one fourth arc tangent. So if you subtract that, that's gonna give you a one fourth. So one half natural log of X squared plus four. We end up with positive one fourth arc tangent of X over two. And then how about this? We have this here. This is gonna be a negative one fourth. Two cancels with that, so it actually gives us a two. So we get X over X squared plus four. There's two down here. I could upgrade this to a four by putting a two down here. And so then combining them, we end up with four minus two X, I'm sorry, just X over two times X squared plus four. And then there's a constant right here. So that was a little bit of a challenge. We were able to work through it. Oops, kind of wrote over my answer there. We were able to get through it. Again, it took a little bit of effort. A trig substitution was involved in this one for that repeated quadratic factor. And that's the main thing I want you to get from this exercise right here is that if you have a repeated quadratic factor that's irreducible, your template has to adjust. So to this thing right here, that your template has to adjust. So you have one for each of the factors. Now in this situation, I kind of did it similar to how we had irreducible quadracks in the previously. But one recommendation I might make to you is that with this thing right here, you might be like, you know what? Partial fraction, decomposition, go away. I'm just going to do a trig sub. I'm going to start off with x equals two tangent theta just on this bad boy right there and go from there. It does turn out to be a lengthy calculation. It's computation intensive. But if you compare it to what we did in this example, it might be that partial fractions was not the most effective way on this one. A pure trig sub would have worked out perfectly for this one. And again, that's kind of leading to what we're going to see in lecture 16 is that as we've learned these different techniques of integration, what's the best one to use when someone's not telling you what to do? And there might be options. This one was successful using partial fractions, but we could also done it from a purely trigometric substitution. And again, it is still a messy one, but it would turn out to have been a little bit cleaner, a teeny bit cleaner. You know, we rolled around the mud for five minutes as opposed to rolling in the mud for 15 minutes. And so that's where we're going to end chapter 15 and think about strategies we can use in the next section. See you then, everyone.