Calculus: Repeated Linear Factors: Part One

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Techniques of Integration
Integration by Partial Fractions with Repeated Linear and Quadratic Factors
Repeated Linear Factors - Part One Page [1 of 2]
All right, let's take a look at some more elaborate rational functions that we want to integrate and see how we can tackle these things.
So here's one that looks really elaborate. Let's look at the integral of dx. I just get tired writing that thing out. Now, if we want to try to evaluate that, how can we do that? Well, it's not obvious how to use a u du-substitution, because what would you do? I guess you would have to let u equal the top, so when you take the derivative you'll have something cute, but then you can't put a du on the bottom. So, in fact, that technique doesn't work. In fact, I don't know what to do with this. However, notice that this thing has the property that the highest degree of the top, which is the x^4 term, is larger than the highest degree on the bottom, which is this 3. So, in fact, long division is going to be a good idea here to first simplify this a little teeny bit. So whenever I see the biggest power on the top exceeding or equaling the biggest power on the bottom, I should first do a polynomial long division. Now, let's see if I can actually fit that on here. These are such big things here. I have to admit, I'll let you in on a secret - I'm a little bit scared.
x^3 - x^2 - x + 1, so there's the bottom part, and then you draw this little curvy thing and then a straight and really long line, because it's long division, and then write this down. Now remember a great technique is to put placeholders in for every single power of x, even if it's not physically there. So here I see x^4, that's good, but I see no x^3. So I write that in with a 0, just to keep things lined up perfectly as placeholders. And I see -2x^2, and then I see a +4x + 1. Okay, and let's long divide. So I look way over here and I say, "What do I have to multiply x^3 by to make it equal x^4?" And the answer is just x, so I put a x here. And now what I do is take x and multiply it all the way through here and write that answer here, just like long division. So I see x^4, and then I see -x^3, and then I see -x^2 and then I see a +x. I put a line under that and then I subtract all of that. So I subtract this and I subtract off this. So what happens? We have to remember to share that negative sign all the way through. So x^4 - x^4, that's 0, and in fact, the first term should always drop out. That's how we determine what to write up here. So that's 0. Here I see 0x^3 -(-x^3), so that's just x^3. Here I see -2^2 minus a minus. That becomes a +x^2, so -2x^2 + x^2 = -1x^2. And then I see a 4x - x = 3x, and then I bring down this term here. And now I repeat. What do I have to multiply x^3 by in order to make it equal x^3? The answer is 1, so I add a 1 next. And then I take 1 and multiply it by everything here, which just requires me to write everything down again. 1 times anything is anything. And then I subtract everything. And when I subtract, what am I left with? Well, the first terms cancel out by construction, and then I see a -x^2, and then -(-x^2), so in fact, they drop out. Here I see a 3x - (-x) = 4x. And then here I see 1 - 1 = 0. So all I'm left with is 4x. So now, let's keep going.
What do I have to multiply x^3 in order to make it equal to 4x? Well, there's no positive powers of x's I can multiply this by in order to make it that. This is actually smaller than this. So just in like the good old-fashioned long division, this must be the remainder. And so, in fact, our long division section of the program here is done. What I see is that this complicated fraction is nothing more than x + 1 plus this remainder divided by this. And so the new integral that we can look at is the following. So now we have a new and improved integral. So the new and improved integral would be, well, this thing is just x + 1 and then + 4x, that's the remainder, all divided by the denominator, x^2 - x^2 - x + 1 dx. So in fact, this complicated integral is actually equal to this complicated integral. Have we made progress? Well, the answer is a little teeny bit, because notice that I can integrate this pretty easily, that's just . I can integrate this pretty easily, that's just x. And now I'm looking at this and you have to admit this is certainly less complicated than this original thing. So I have, at least, I'm biting into the question. I'm biting into it little by little.
Now, I look at this piece and I see what could I do? Well, I don't know. It looks pretty awful to me, so what do you do? Well, if you look at this and just look for patterns wherever you can find them, what I do notice is that in these first two terms I could factor our an x^2. Now, if I did that, what would I be left with? So let's just try that and see what happens. If I factor out x^2 down there, I'd be left with x^2, I have a x, and then I have a -1. And then I have this stuff here, -x + 1. I'm just looking at the denominator right now, just the bottom. Okay, well that doesn't look too helpful, except you may notice that this is x - 1 and this is almost x - 1 here. In fact, if I factor out a negative sign, what would I see? Well, if I factor out just one negative sign, I would see x - 1. Let's just check that, because that's -x and a negative and a negative is a positive 1. So, in fact, I see that this denominator is actually nothing more than x^2(x - 1) - (x -1). Well, now I see another common factor. I see x - 1 as a factor in this piece, but also x - 1 is a factor in this piece. So I can factor that whole thing out now. Now what happens if I factor that whole thing out? If I factor that whole thing out, what am I left with? Well, in this term I'm just left with the x^2, and in this term I'm just left with that negative sign, -1. So, in fact, this denominator, it turns out, can be factored. Now, you sometimes see these factorizations, why you've got to look at these things sort of in many different directions - first of all, if you could factor it by looking at it, that's great. If you can't, it's not a bad idea every once in a while to try to just factor a little piece of it and then try to find a commonality and then factor more stuff out. In this case, when you use that technique, I actually see that the bottom can be factored quite nicely. In fact, this can be factored, too. This is the difference of two perfect squares. So this is actually (x + 1)(x - 1). So, in fact, if I just do some more stuff of this sort, what I see is x - 1, there's that first piece, and this factors into (x - 1)(x + 1). And I can combine these two, since they're the same now. And so I just see (x - 1)^2(x + 1).
So the moral of the story so far is that just by doing all this little bit of arithmetic and long division, I was able to take this really complicated problem and convert it into a problem that now looks like what? Well, the new integral, and maybe you're not satisfied, maybe you're saying, "Gee, it still looks pretty awful." Well, okay, let's see what happens. This would now be the integral x + 1, don't change anything up there, plus - and then the only thing we did up here is just to factor the bottom so I have the 4x on top, and now the bottom I see is just (x - 1)^2(x + 1) dx.
So the first step in resolving this original integral was to realize I needed to do some long division. So I did some long division and I see that I can rewrite it in this form. So now I've got some easy terms and this term seems a little bit more simple to deal with than the original thing, so that's good. And then I try to factor the bottom and after factoring the bottom, we now realize that the original integral is nothing more than this integral. Now, this actually is a little bit of progress, because these two things we can certainly integrate, and now this term right here I can use the technique of partial fractions to rip this apart into many pieces. Now, there's a little teeny trick or intrigue here, and that is I see that this thing is occurring to a second power. It's a square. So how do I deal with something that's actually appearing with a square? So you have some quantity squared. I turns out we're going to have to modify ever so slightly our partial fraction technique to capture that. I'll show you that in the next lecture. I'll see you there.