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Calculus: Distinct and Repeated Quadratic Factors

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  • Type: Video Tutorial
  • Length: 11:09
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 120 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
Distinct and Repeated Quadratic Factors Page [1 of 3]
Okay, so let me just try to recap what we've seen when we have repeated linear factors. So suppose you want to do the partial fraction decomposition for this particular rational thing. Now you can see what I have here is a linear factor squared, another linear factor cubed and just one linear factor all by the person's self. So how would you break this up? Well, remember that you want to give credit for every single factor every time that thing appears. So what I would do here is set this up as follows: you would say this equals , just have it sort of naked, and then plus . Don't forget the square. And then plus , and then plus , and then another one, . So for every power, you basically write down a representative for that power. And then finally, don't forget the one that's all by himself, that's . And now you get a huge common denominator there and you put it all together and so forth and set all the equations together and do all that. And if you do that it turns out that, in this problem, this is what the answer is. So the A = , the B = . You can see it gets pretty complicated, but now if this were an integration problem, you would just have to look at these things individually. So if you see higher powers of linear things, then you're fine.
But what I want to share with you now is the thinking about what happens if you, in fact, don't have a power of a linear thing, but you just have a quadratic by itself. Not jut a linear thing to a power, but you see something that's a quadratic that's not a linear thing to a power.
So let me just go right to an example so you can see what I'm talking about. So here's an example. Let's find the integral of . Okay, so let's immediately jump into this thing. I see that the degree of the bottom is bigger than the degree of the top, so I don't have to long divide, thank goodness. So what should I try to do? By the way, a u-substitution's not going to work here if we try it. So the only technique that's sort of in town here is the partial fraction technique. Let's see if the bottom can be factored. So let's first of all forget about the calculus problem and just look at the intergrand. So can the bottom be factored? Well, let's see. The bottom, in fact, I can factor out an x^2, and then I'd see x^2 + 1. So the bottom actually can be factored into these two factors. This can't be factored anymore over real numbers. That's not the difference of two perfect squares or anything. So, in fact, all I can say about this person right here is that it equals, well, the top - it's all that stuff there, divided by x^2 (x^2 + 1). And the way I look at this and the way I want you to look at this is to realize that these are two kind of different-looking things. This is just an x and that's just a linear thing, squared. So we'll have to use the same method that we used before. But this is genuinely a quadratic thing, because it's not just a linear thing with a big square on the outside. This is a quadratic that can't be factored anymore. So we need to treat this as one separate new object. So how do we deal with quadratics? Well, let's think about how we dealt with the linear stuff to break it up into partial fractions. What I'd do is I'd say we'll I'll just put , and then to pay homage to the fact that it's actually appearing twice, I put . But now for this quadratic factor, I actually will up the ante on the top. See, when this thing was just an x, I put just a constant number here. If this thing is a quadratic, I'm going to put a linear object here. So when I see a quadratic thing like this, what I put is Cx + D. I put a general linear thing here. When I've got just a linear thing here, I put a sort of one degree less than a linear thing, would be a constant thing. So I just put a constant. If I have a quadratic, I put sort of one degree less, which would be just a linear thing. So when you see a quadratic that you can't factor anymore, then what you want to do is put on top not just a constant, but to pay homage to the fact that this is really a quadratic thing and put Cx + D. Not to be confused with when you have a linear thing to a power, when you just repeat that thing as many times as you need to.
Okay, well now you've got to get a common denominator and do all this stuff and figure it out. But this is sort of like one of those cooking shows - you know those cooking shows, where you say, "Well, okay, and to bake the cake, you have your frosting and you pour it right in, and you spread it around." A sort of Julia Child of mathematics here. And you put it in, nice and round, and make sure it's spread out evenly, let it sit a little bit, and then bake this for 4 hours at 650 degrees. And if you wait 4 hours - oh, but I happen to have one already done. You get this. So that's the cooking show version, and now how does the math version go? The math version would go like this: if you do all that out, what you see is that this thing is equal to this thing. So you can actually figure out what the A should be, what the B should be, the C and the D. A is going to be 2, B is going to be -1, C is going to be 3 and D is going to be -2. But the moral of the story is that when you see a quadratic that can't be factored, the thing you put on top is not a constant, but, in fact, will be a linear thing. So it'd be Cx + D as the general linear-looking thing.
So to find this integral, so it's the recap here, to evaluate this integral, all I've got to do is evaluate this integral. Okay, and that integral, well, that's going to require a teeny bit of work, but not too much work. Let's see if we can get through it. This is going to be not too bad, this is going to be not too bad, but this looks a little bit scary. You could try a u-substitution, but that's actually not going to quite fly here. So I've got to actually break this up, and so let's just focus on that little piece of it right now. So if we focus on that piece, what I see is dx. So I'm going to actually break this apart into two fractions. And so one fraction looks like this, , and the other fraction is . I can always do that as long as I keep the same bottom there. If you just combine these things, you get back to here, 3x - 2. So these though can be integrated separately. It is just a u du-substitution and this actually is one of these new functions that maybe we're just getting familiar with. This is going to be an arctan. So let's see how this would sort of play out. So if we put this here now, what I see is, see if we can report the news here, this would equal - so this integral is going to be just , and then what's this integral here? Let's do that as a little side calculation. We're here, we're integrating . That equals -x^-2, and so what is that? If I add 1, I see , so this is just . So this is just going to be , so plus . And then what about this term here? Well that term we broke up into these two separate terms. And this term is going to be another little u du-substitution. And if I let u equal - so let me just write this down really quickly here. So I have + 1 dx. Let's let u = x^2 + 1 and then du = 2x dx. I have x dx, so I'd better divide by sides by 2. And so I see this new integral becomes = and just the integral of . So it's a natural log. So it's , which is x^2 + 1. So if I write that in, I see this piece right here, which part of this is plus . And then this final term, -2, and that's just the arctan of x, one of the ones we saw earlier.
I want to just make a little comment here, by the way. You don't actually need the absolute values here, since if you take x and square it, that can never be negative, and if I add 1, in fact, this will always be positive. So, in fact, the absolute values are a little bit superfluous, but absolutely correct. But you don't need them here.
Anyway, the original question was to evaluate this integral and we saw that, after doing an appropriate partial fraction decomposition, we could actually convert it to this integral, and that integral can be solved. Again the moral is that whenever you see a quadratic down here that can't be factored when you set up the partial fraction decomposition, make sure that you put a Cx + D, a linear factor, on top. If you have just a linear thing here, then just put a constant number on top. So that's the moral of the story.
What about if you have repeated quadratics? Well, repeated quadratics would be treated just like you treat repeated linear stuff. For example, let's just decompose . How would you break that up? Well, I see a quadratic appearing twice. So you know how I break that up? I say, well, first of all, I've got this , so that's just here, and then I've got to remember the square. So I put , and then another copy, so . And then if you solve all that, you can actually find the coefficients. This is , this is , this turns out to be 0, this turn out to be , and this turns out to be 1. So again, just like the cooking shows, I did it for you in advance, but this would actually require a lot of algebra.
So when you are thinking about these partial fractions things, I want you to think about whether you have a linear factor or a quadratic factor. And if you have a linear factor repeated, don't forget the square and put it all down. And if you have a quadratic factor, remember on top you want to have a Cx + D. Bon apetite. It tastes a little stale. I think I let it sit too long to cool off, but delicious. Enjoy.

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