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Calculus: Repeated Linear Factors: Part Two

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  • Type: Video Tutorial
  • Length: 15:20
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 165 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

Calculus (279 lessons, $198.00)
Calculus: Techniques of Integration (28 lessons, $40.59)
Calculus: Integration: Partial Fractions, Division (7 lessons, $11.88)

Taught by Professor Edward Burger, this lesson comes from a comprehensive Calculus course. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/calculus. The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hopital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, "Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas" and of the textbook "The Heart of Mathematics: An Invitation to Effective Thinking". He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The "Journal of Number Theory" and "American Mathematical Monthly". His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

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Techniques of Integration
Integration by Partial Fractions with Repeated Linear and Quadratic Factors
Repeated Linear Fractions - Part Two Page [1 of 3]
All right, so I want to try to figure out what the anti-derivative is of this really complicated ration function, this ratio of polynomial divided by polynomial. The first thing I realized was that the degree of the top is actually greater than or equal to the degree of the bottom, so I actually did a long division and converted the question to a slightly easier question, namely, this one. Well, I don't know how easy it looks, but I can integrate that, I can integrate that, and now I'm left with this thing. So that thing inspires me to think about trying to break that up into partial fractions. So the idea is if I see something like this, what I really want to do is break it up and say - you see something like this, I want to really break it up and say, "Well, actually it came from two different pieces." So if I see this factorization, like I do here, I want to split it up into two different pieces and then try to integrate each of these separately, which might be a lot easier. So the method really here is to divide and conquer.
So what I want to do is take a look at this piece right here and see if we can break that down into two partial fractions. So what happens when we try? So let's give this a try. So we've got, well, we've got this, dx. So that's what I want to break up. So how am I going to break that up? Well, let me break that up in the way that we seem to be thinking about these things. So I would take - well, I don't know, how would I do it? Would I put - see, how do I deal with this square? See, in fact, that's sort of an annoying feature. How do you deal with that square? When I see something like this, just an x to the first power here, but the whole thing is squared. What should I do? Well, what I've got to do is to make sure that I get the square in. Well, you may say, "Well, gee, all you've got to do is write in something like and then plus . Or you may say, "Well, gee, maybe I should write that as (x - 1) (x - 1), and then put two copies, + + . Well, what I want to do is sort of combine these ideas and what I want to call your attention to is whenever you see something like this, where it's all squared, you see x to the first power with stuff all square, you've got to remember and remember not to forget the square part. So here's how you resolve that using partial fractions. What we do is I write , just like before, and then I don't forget the square, so I now add in a term just to pay homage to the square. So I put in one of these. So when I see something like this, x to the first power times stuff squared, what I do is I put in two things; one with just the thing naked and then one with that thing squared. And then don't forget the last term, and that last term just appears on its own, so I just give it its usual billing, and I see that. So whenever you see a linear-type thing raised to the square power, what you do is you first write A over the linear thing and then plus B over that linear thing squared, and then plus whatever other terms you have. And that's the way you resolve this paradox about having an exponent of 2 there.
Okay, now let's just use the partial fractions technique and see where that gets us. So what do I do? What I have to do is I have to get a big common denominator and then set this whole thing equal to that. So let's get that big common denominator right now and see what this looks like. So what would the common denominator look like? Well, let's see, here I'd have to multiply top and bottom by x - 1 to make sure I have an (x - 1)^2 over here downstairs, but I also have to multiply top and bottom here by the x + 1, in order to make sure that I've got that term represented in here. So I have to multiply top and bottom here by (x - 1)(x + 1). So that's a mouthful, but let me write that down for you. So I'd have A(x - 1)(x + 1), and that's all over this now new common bottom, (x - 1)^2 (x + 1). And you can always check to make sure that this quantity genuinely equals this, because notice that these two terms just cancel away. And notice that this term cancels away with one of these and it just leaves me with , which is what I have here. So you can always check, but now I've got that nice common bottom, plus - and then what do I have to do with the B top? Well, I already have the (x - 1)^2 stuff, so that's good. All I need is the x + 1, so I multiply top and bottom here by x + 1. So B(x + 1) all over the common bottom (x - 1)^2 (x + 1). And then finally, I've got this last term. What do I need here? Well, I just need those two copies of x - 1. So I've got to multiply top and bottom by (x - 1)^2. So I have C(x - 1)^2 all over the bottom (x - 1)^2 (x + 1). So now everything is over a common bottom, and so I can actually just combine by adding the tops. So if I do that, let's see what we get. By the way, if you FOIL it out, you'd see x^2 - 1 + B x + 1 + C, and maybe I should just square this thing out. So if you square that out, I think we see x^2 - 2x + 1. And that's all divided by the common bottom. So I'll write that as (x - 1)^2 (x + 1). By the way, with these partial fraction things, you can certainly see that these questions and these algebraic things get really long, so two pieces of advice: number one, write very neatly and two, write very large. If you write small, there's a easy tendency to make mistakes. If you write really big, it's less likely, but even then it's hard to avoid making some little arithmetic mistakes.
Okay, now I take a look at this and what I want to do is I want to combine all the like terms in terms of the x's. So, for example, if I were to distribute this, I would see Ax^2 - A. Here I'd see Bx + B. Here I'd see Cx^2 -2Cx + C. So I want to combine all the different types of x's together. So let's pull together all the x^2 terms. I see there's two of them here, so that would be, there's an A, Ax^2, and there's also this term right here, that's a C. So A + C is the coefficient on the x^2 terms, factoring that out. What about the x terms? Well, I see a Bx here and I see a -2Cx here. So, in fact, I could write that as (B - 2C)x. Because if you notice, there's a -2Cx here and there's a Bx here, so I factor those x's out and I get this. And then lastly, what about my constant terms? Are there any constant terms? Sure there are. There's a -A right here, don't forget that minus sign when I distribute, there's a B, and then there's a C. By the way, what you might want to do, if you have a lot more room than I do here, is to actually write this out and just distribute everything. Write Ax^2 - A + Bx + B + Cx^2 - 2Cx + C, and then you can really just move the terms around. But I'm running out of space here, and so I really want to get this down. So I'm just trying to condense it just a teeny bit, but it's the same exact idea. And it's still over the common bottom. So the common bottom, which is (x - 1)^2 (x + 1).
Now, what's the point of all this? Well, the point of all this is that all of this hard work was to take this thing, break it up and then to see what it equals. And what I want this to do is I want this to actually equal that thing, so let' see if we can make that equal that thing. If I make that equal it, I want this to be the same as this. So I want these to be equal. And so what I really want to think about this is I want to think about it as an identity. It's really an identity. That means that since the bottom is exactly the same as the bottom here, I want the top here to be exactly the same as the top here, exactly the same. So what does that mean? Well, it means that we can say a whole bunch of things about these terms. For example, notice there's no x^2's here at all. That means that this term must be 0, because I don't want any x^2's here. So therefore that tells me that A + C = 0. Now, do I have any x's? Yeah, I've got four of them, so this term in front of the x must be 4. So I have B - 2C = 4. And finally, do I have any constant number, like +3 or +7? No, there's no constant term, so all that constant stuff must equal 0. So I have -A + B + C = 0. So now I have three different equations and I have three different unknowns.
And so now I've got to figure out how to solve that for the different unknowns. And the way that I would do this is just to look at these and try to get rid of some of these unknowns. For example, I can take this equation here and solve it for one or the other. By the way, there's no just one way of doing this. In fact, there are tons of different ways of doing it. I'm just going to show you one way right now live. So one way is to take this first equation and solve it for A, and I'd see that A, using the first equation, would equal -C. Now, if I take that value A = -C, I could actually go into here, where I see an A, and insert the value -C. And if I do that, what would I get? So now I'm going to insert this into that equation and what would I see? Well, I would see that in place of A, I'm going to put -C, so I see negative in front, so -(-C) is just a C. So I see C + B + C = 0. So that means that B + 2C = 0. So if B + 2C = 0, then what do I do now? Well, I haven't used this equation yet, so now let me take a look at this equation. I could actually now solve this for B. And if I solve this for B, I see that B = -2C. I just took this and brought this over. But look, -2C, there it is. So in place of this -2C, I could insert its twin, namely, B. So this is just B + B, because the -2C is replaced by just a B. And so I see B + B = 2B = 4. And so therefore if 2B = 4, the B = 2. I just found the B value. But now that cracks the whole thing open, because if I know B, I can find everything. You see, if B = 2, then I can go back to here and I can figure out what C would have to B. Because if B = 2, then that means that 2 - I'm using this equation right here, 2 = -2C. Well that would mean that C = -1. So now I know what C is. So C = -1, and then using that, I can come back here and figure out what A is. A would have to be -(-1), A = 1.
Now there are many ways of solving these three equations and three unknowns, but the substitution way really works, and you can substitute any different way you want. Anyway, the moral is we now know what those A, B, C and so forth really are. So if I come back to here - a long time ago we had this, what I see now is that this can be expressed as - oh my goodness, wait a minute, don't go anywhere - like this. And now I know what the A, B and C are. What are the A, B and C? So the A= 1, so this is now 1, the B = 2, and the C = -1. And so, in fact, what I see is that I can rewrite that integral. That integral can now be rewritten as the integral x + 1 - and now in place of this, I could write dx. So this complicated integral now becomes this, and now this we could integrate just term by term. And so what does this equal? Well, the integral of x = . The integral of 1 = x. What's the integral of this? It's . You might want to do a little u du-substitution to verify that. What about this term right here? Well that looks a little bit threatening, but that's just another u du-substitution, and so let me just look at that little piece of the integral right there. I see . How would you integrate that? If I let u = x - 1, then du = dx, and so really I'm just integrating du. And so what's that? That's just the integral of u^-2 du, and so that equals , and so that equals . And what's u? It's x - 1. So that little integral right there just turns out to be that. Don't forget the 2 here in front, so what I see here is a . That's that little integral right there. And then this term finally, it's another + C.
So, the moral of the story really is that this complicated integral can be integrated. We can actually find the anti-derivative of this complicated function, first with the long division, then with an application of partial fractions to break it up like this, and then notice each and every term is actually not too hard to break up. Divide and conquer partial fractions. And when you see a linear thing squared, don't forget to write out the linear thing and then plus the linear thing squared to make your decomposition. I'll see you at the next lecture.

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