 In this example, let's evaluate the integral of the function x over x squared minus 5x plus six with respect to x here. This is a rational function. So in order to find the anti-derivative, we're gonna probably gonna have to use the technique of partial fraction decomposition. We wanna rewrite this rational function as a sum of polynomials or partial fractions, proper fractions. Proper meaning here that we want the numerator to have smaller degree than the denominator. In this situation, the numerator has a degree one, the denominator has a degree two, so this is a proper fraction. So we can skip any need to use polynomial division of any kind. So the next thing we need to do is we're gonna have to decompose this rational function into a sum of partial fractions, each of which will be of a form that we can integrate, because that's our goal. We wanna decompose this thing so we can integrate this. And decompose here, it doesn't mean like there's some dead raccoon on the side of the highway. What of course it means is we wanna break this up into smaller, simpler pieces. So with this decomposition, step one, what we need to do is we need to factor the denominator. It's a crime when someone leaves their denominator unfactored like in the situation, but we can live with it. We need to factor the denominator, in which case here that is we're factoring x squared minus five x plus six. This situation's not so horrible. We have a quadratic trinomial whose leading coefficient is one. So we're looking for factors of six that'll add up to be negative five. So we can take negative two and negative three. Notice if you were to FOIL us out, x times x is x squared, negative two times negative three is a positive six. And then if we get negative three x and negative two x, that adds up to be negative five x. We found the correct factorization, this we need to know, all right? The next thing we need to do is we need to find the template, the template for our partial fraction decomposition. Now, back in the early days of Microsoft Word, if you were typing something on your computer, like dear Bartholomew, there was a little animation of a paper clip, clippy, who would appear and tell us things like, oh, it appears that you're trying to write a letter or oh, it appears you're trying to write a resume. I can offer you some templates, which a template means that we want something that's already pre-structured, right? The formatting is done for us. We just have to sort of enter our specifics. Like if we were writing our resume, we would type in our name, our experience, work experience, our skills, things like that. So for partial fraction decompositions or PFDs for short, they have templates as well. And the templates can be based upon the factorization of the denominator. So for us, our template X over X squared minus 5X plus 6, the template's gonna have the following. Remember, we're trying to decompose this into fractions, which will be any easier to integrate. And we're gonna get two fractions that coincide with the two factors of the denominator we found before. So the first fraction will have as its denominator X minus two and the second will have as its denominator X minus three. But what the numerators are, we'll come back to those in just a second. So let me kind of explain why it's gonna break up like this. Because if we were to add two fractions together, if the first fraction's denominator was X minus one and the second denominator's fraction was X minus three, if we add these together, we'd have to first find a common denominator. The least common denominator would be their product, X minus two and X minus three. And then we could add these things together. So if this fraction right here is a sum of two partial fractions, it must have been the case that one of the fraction had a denominator of the first factor and the other fraction had a denominator of the second factor. So we can add those things together. But what are the numerators gonna look like? Well, this is the idea of why it's important that this fraction is a proper fraction. If we have these two, if we have a proper fraction right here, then we can assume it's a sum of two proper fractions themselves, that is something less than one, right? So by adding together two proper fractions, we get a proper fraction. Now if the first denominator is X minus two, if it's a proper fraction, that means its numerator has to be a degree less than one, well the only option there would be it's a degree zero polynomial, which is a constant. So we actually know there's some number on top. We'll call it capital A for the moment. We get some capital A and then for the other one, there's some other capital, which we'll call it B. And so this right here is our template based upon the factorization X over X squared minus, X over X squared minus five X plus six. It'll break up as two partial fractions, something over X minus two and something over X minus three where those somethings are gonna be real numbers. All right, so now we have our template. We're pretty good. Now that you look at the template, the template itself is an equation. And so the step three here is you're gonna clear the denominators in this equation above, clear the denominators. That is you're gonna multiply both sides of the equation by the LCD. So on the left hand side, you multiply by the LCD. On the right hand side, you're also gonna multiply by the LCD. Now on the left hand side, the least common denominator is of course, just X squared minus five X plus six, that is X minus two times X minus three. So they'll cancel out indiscriminately. On the right hand side, you're gonna have to distribute this thing through. So you end up with X is equal to A over X minus two times by X minus two and X minus three. And then with the second one, you're gonna get B over X minus three. You times that by X minus two and X minus three. And there should be some cancellation that happens here. For the first fraction, you notice that X minus two cancels out with the X minus two in the bottom. And for the second partial fraction, the X minus three's cancel. And so doing this, our simplified equation won't have any more fractions. It looks like X equals A times X minus three plus B times X minus two. So all of the fractions are now gone. And that's to say we've simplified and we've simplified by clearing out the denominators in a manner similar to this, right? So we get X equals A X minus three plus B times X minus two, all right? So the next thing we wanna do is we wanna set up a system of equations, system of linear equations. So set up a linear system. So how does one go about doing that? What we're gonna see is the following. What I want you to do is take these coefficients A and B and distribute them through here. And if you do that, you end up with X equals AX minus three A plus BX minus two B. Now looking at this equation, I want you to be aware that we have some variable X and it's a variable that means its number is allowed to vary. X could be one, it could be two, it could be three. It could be lots of different assignments, right? Now the numbers A and B aren't variables in fact. I mean, they're unknowns, but they're not variables meaning they're not allowed to vary. There's only one choice of A that's gonna work. There's only one choice of B that's gonna work. We just don't know what they are right now and we're using the symbols A and B as placeholders for what those are gonna be. And so the reason I mentioned that distinction here is if we try to combine like terms on the right-hand side, X is our variable. We add together things like AX plus BX. So how do you add together AX plus BX? We'll just add them together. A plus B times X. Now we don't know how to simplify A plus B yet because we don't know what A, B and R but we can add together the coefficients of X. And we can also add together the constant terms. Negative three A and negative two B don't involve X whatsoever. So this would be considered a constant. Negative three A minus two B. This is what the right-hand side is gonna look like. The left-hand side is much simpler. You end up with one X plus zero. Now you don't typically write the one and the zero here because it's implicit but the reason I wanna do that here is to make a connection. That when you look at the right-hand side, the coefficient of X is A plus B. But on the left-hand side, the coefficient of X is a one. And the only way these two polynomials could be equal to each other is if their coefficients are equal. We have to equate these things together. And so we actually get the statement that A plus B, the coefficient on the right, is equal to one, the coefficient on the left. Now we can also do this for the constant term as well. On the right-hand side, the constant term, that is that term that doesn't involve any X's whatsoever, is a negative three A minus two B. On the right-hand side, it's a zero. And so the only way that these two polynomials could be equal is if these are equated as well. In which case we didn't get negative three A minus two B equals zero. And this now gives us a system of two linear equations with two unknowns. And there's a couple ways one could try to solve this system. We could use elimination, substitution, we could use matrices, determinants. There's a couple of ways and many of you might have seen this in the past, but that actually gives us the next step in our process here. Since we are on step four. So now step five here is we want to solve said system of equation. And again, there's a lot of ways that one could go about doing this. I'm gonna solve it just by elimination. So I'm gonna take the first equation times everything by negative two. Upon doing that, your system of equation will convert. You'll get negative two A minus two B. I'm sorry, I wanna do a plus two. So we're gonna get plus two A plus two B is equal to two. And then the other equation is negative three A minus two B equals zero. The reason I wanna have a plus two is I want that when you combine the terms that one of the coefficients is opposite but equal. So we have a two B and a negative two B. So when you add those together, the B's will cancel. The two A plus negative three A gives you a negative A and then you get two plus zero, which is two. Solving for A, we then get that A is equal to negative two. Like so. And now that we know that A is negative two, we can actually plug that into our previous equation and solve for B. We get that negative two plus B is equal to one. So add two to both sides. We see that B equals three, one plus two. And so with that in mind, we have A and B. Remember, these were coefficients of our template. Coming back above here, we have this template right here. So we can now substitute the numbers A and B in what we found here. So I'm gonna just put that down here. So remember, we're trying to calculate the integral X over X squared minus five X plus six DX by this technique of partial fraction decomposition. We have now discovered that the fraction can be rewritten as three over X minus three minus two over X minus two. I switched the order and I wanna put the negative one second. But this is the function that we want to integrate now. And it's what we've seen already just by a basic U substitution. You have three over X minus three here. You could bring the three out. You're left with one over X minus three. This is where the U substitution comes into play. The antiderivative of one over X minus three will be the natural log of X minus three. And so a very quick calculation, we get three times the natural log of the absolute value of X minus three. We're gonna get minus two times the natural log of the absolute value of X minus two plus a constant. And this is our antiderivative here. Now, I wanna emphasize that in this example, the actual calculus that we did was fairly simple. Just using the U substitutions here. We've done this before, which is why I kinda skirted through this one pretty quickly. The calculus is actually pretty easy right here. This whole shebang came about from a very lengthy algebraic calculation. It took us a while to find this decomposition. Took a while to find the A and the B, all right? But once we found it, we could convert the rational function to this partial fraction decomposition and then find the antiderivative very quickly. So we need the PFD to help us find the antiderivative, but the PFD is the hard part of these problems, finding the partial fraction decomposition. It involves this type of decomposition process using the system of linear equations. If it's an improper fraction, it might involve polynomial division. And so these things are quite lengthy, yes, but it's a helpful algebraic tool to solve these type of problems. Now, in the next video, I wanna do another one very similar to this, but I wanna show you a different algebraic technique. If you're not a big fan of systems of equations, there is another technique one could use. So stay tuned for that.