 Hello, and welcome to the screencast on constructing a partial fraction decomposition. Today we're going to integrate this function. In fact, this function is a particular kind of function. It's a rational function. That is, it's a polynomial divided by another polynomial. This is the only kind of situation where we can use partial fractions. Before we do that, it's worth trying other possibilities. In particular, substitutions often work on fractions. Pause the video, take a moment, and try to integrate this using a substitution, and see what happens. Alright, we're back. Hopefully you discovered that if you choose u to be the denominator, you can't quite make u appear or du appear in the numerator. It's close, but not quite there. And that's really the only way we could have done a substitution. So we are left with partial fractions. Something else that's important to remember about partial fractions. It is not an integration technique, so much as it is an algebraic technique. It's a way of taking a fraction and splitting it apart into simpler pieces that we can hopefully integrate more easily. That means that to start partial fractions, I'm going to leave the integral sign off entirely and just recopy the inside here. And we're going to just rewrite that inside bit, hopefully as simpler fractions. The first step is to factor the denominator so that we know what those simpler fractions are going to look like. So 4x plus 2 won't change. We're only going to focus on the denominator. And it turns out that this denominator factors into an x plus 5 and an x minus 1. Make sure you know where that comes from. You can always reverse engineer by multiplying those factors back out and making sure that you get the right thing. And now we're going to write the prototype for the partial fractions decomposition. We're going to write what it will look like with a few unknowns in it. So our fractions are going to have each of these factors. We're going to have one fraction per factor. So there will be a fraction for an x plus 5. And there will be a fraction for an x minus 1. But we don't know what the numerators are. And so we're going to call them constants a and b. And our goal is to solve for a and b. If we do, we'll have split this fraction apart into these simpler pieces that we know how to integrate. Our goal is to solve for a and b. That's the purpose of everything else that we're going to be doing right now. Keep that in mind. So in order to do that, we're going to start by doing a bit of algebra to get rid of these extra fractions that we have sitting around. So I'm going to rewrite just the key parts of this. We have the fraction 4x plus 2. So the original fraction. And then leaving a bit of space for what's about to happen, I'm going to rewrite each of these on its own. And the reason that I'm leaving so much space and moving this one over a little bit is that we're going to clear the fraction by multiplying through by the denominator of the big fraction of the original one. We can multiply both sides by x plus 5 times x minus 1. Because we've factored this in such a way that every fraction has at least one of those factors in it, this guarantees that something's going to cancel in every single fraction. So this is really just an algebraic move. It's true that we can multiply through, and even better, a whole lot of things cancel on the left and a whole lot of things cancel on the right as well. Rewriting what we're left with, I have only a 4x plus 2 on this side, and I have only an a times x minus 1 here and only a b times x plus 5 there. We've successfully gotten rid of the entire set of fractions, and now we have just polynomials. Keep in mind, we're trying to solve for a and b. So now we're at a point where things are much simpler and we can start to actually find values for a and b. There's a number of different ways that we can do this, but the easiest thing to do is to try some convenient x values, which will make parts of this formula disappear. So notice we have three things we don't know, a, b, and x. We only want a and b. x is a variable and we can leave it there. And so I'm going to choose a value such as x equals 1, which causes both a and an x to disappear. So if I plug in x equals 1, I'll get a 6 on this side, and on the right hand side I get a times 0 plus b times 6, which turns into 6 equals 6b. And so I've managed to find that b equals 1. By choosing the right value for x here, I cleverly managed to make the a disappear and that let me solve for b. So take a moment, pause the video, and think, what would you do in order to solve for a? Alright, in order to solve for a, I want to make the b term disappear. So I'm going to choose x equals minus 5, which will turn that term into 0. When I plug in minus 5, the left hand side becomes minus 18, the a term gets a minus 6 coefficient, and the b term gets a 0. So I have minus 18 equals minus 6a, and so a is 3. Take a look back at that algebra and make sure you know why each of those steps happened. We're finally at the point where we can rewrite the original big fraction in terms of these smaller fractions because we found each of the unknown parts, a and b. So rewriting the original, the thing that we wanted to start with, this is equal to a over x plus 5. Well, now I know what a is. It's 3 over x plus 5, and now I have b over x minus 1, but I know what b is. 1 over x minus 1. And that has allowed me to break this fraction apart into simpler pieces. Now, everything that you see on the screen right now was just algebra to simplify the fraction. We haven't integrated yet. So on the next screen, we're actually going to do this integration. All right, I've rewritten what we found on the last screen. In order to do the integral, things are relatively quick now. We wanted the integral of this big fraction, but now I know how to rewrite that. That's the same as the integral of these two smaller fractions. And I know that I can break that integral apart into two pieces, each of which is a simple fraction that I can do by substitution. So now each of these really needs to be done by a substitution, but that substitution is very simple. So here I'm going to actually write out what the substitution is for the first one. If I set u to be x plus 5, then du is just dx. Simultaneously, I'm going to do a substitution for this right-hand integral. I'll call it w. And again, I choose the denominator so that dw is dx, a very simple value. So that means that when I actually rewrite these, I get the integral of 3 over u du plus the integral of 1 over w dw. And I can integrate those each pretty easily. So this is 3 times the natural log of the absolute value of u plus 1 times the natural log of the absolute value of w plus an unknown constant. And then finally, rewriting everything in terms of u and w, in terms of x now, I get 3 times the natural log of, and I know what u is. It's x plus 5. And I know what w is. It's x minus 1 plus an unknown constant. This was a lot of steps. This is the only place where we actually did calculus. We used the partial fractions decomposition in simpler fractions, which we could then integrate using a substitution. And you'll eventually get to the point where you can see those substitutions and do them right away. They each turned out to be logs, which almost always happens when you have a partial fractions decomposition. Take a look back through all of this. Make sure you know what's going on. And remember that in the end, this was really about algebra and about how to break apart fractions. The calculus was things that you already knew.