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Calculus: Partial Fractions


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About this Lesson

  • Type: Video Tutorial
  • Length: 10:58
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 118 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 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
Introduction to Integration by Partial Fractions
Partial Fractions Page [1 of 2]
Okay, let's say more about this idea of partial fractions, where we have an integral, where we can factor the denominator of the integral and then the idea of taking one fraction and breaking it apart into a whole bunch of little fractions, whose sum or difference is the big thing. So here's an integral that I want us to take a look at next.
The integral of . Now happily here in this question, the bottom is already factored, so we don't have to realize that, in fact, the bottom is factored. It's given to us. But if it was something sort of unfactored, we might want to try to see if we could actually factor that quadratic thing and see if we could get two linear things. But in this case, it's already factored. So let's see if we can use the technique of partial fractions to take this fraction, whose integral seems a little bit complicated - how would you tackle that - and see if we can break it up into two fractions, each of which we can actually integrate. So now we're going to spend some time just looking at that fraction and trying to deconstruct it.
So if we try to deconstruct that fraction, how do we proceed? We take . And I'm going to set that equal to . And so now our mission, the first step in conquering this integral, is to just figure out what numbers should go here. So, in fact, this is really going to be an identity. I want to say, "It's an identity." The question is what should A and B be? Well, the way to do that is just to take this fraction and now push forward, get a common denominator and figure out what this is one big fraction, and then force that fraction to look physically like this. So let's take that on as a challenge.
So the first thing we'll do is just try to get a common denominator. That requires me to multiply top and bottom of this fraction by x - 2 and top and bottom of this fraction by 2x + 1. And if we do that, what do we see? Well, I'll see. And you'll notice here that these terms cancel and I'm just left with , just like before, and here these two terms cancel, I'm left with . So I did nothing but get a common bottom. However, now I'm allowed to now combine this whole thing and I'm going to write everything just over this common bottom. But, concurrently, let me distribute the A, multiply this number through by the x, and also the -2, and similarly here with the B. So what I think I see at the end is . Okay, now let's try to factor together like terms and the like terms that I'm thinking about are I want all the x's together on top and I want all the constants together on top. So I see some x's here, so let me actually combine these things, and what I see here is (A + 2B)x. So those are all the x terms. And then what about the constant terms? Well, that would be + B - 2A. So those are all the constant terms, and then it's still over that same crazy bottom, (2x + 1)(x - 2). And so, if you take these two fractions and you get a common denominator, it becomes this. Now, look, I'm trying to find the A and the B so that this fraction is identical to this one. That's the point, to figure out what number I should put here, what number I should put here so these two things are identical. So we visually look at this and say, "What's required to make these things identical?" Well, there are no x's here, and so this thing multiplying the x must be 0. I don't want the x there. So that gives me the fact that A + 2B = 0. And then what about this number? Well that number has to correspond to the number I see here, and so therefore I must have that B - 2A = 5. Okay, so that's where I am right now. What I see is I have A + 2B = 0 and, simultaneously, B - 2A = 5. And now I can solve these things by, for example, taking this thing and solving this for A. If I take this equation and solve it for A, I see that A = -2B, just bring this 2B over to that side. And now I can take that value for A and stick it in here. So if I insert this value for A into this thing, I see B, and then I have -2A, so -2(-2B). That's now 4B = 5. And that means that 5B = 5, which means that B = 1. So if B = 1, what does A have to equal? Well, you can just go right back up to here and see that if B = 1, then A = -2. So now I see that, in fact, these mystery numbers are understood. B = 1, so this has to be 1, and A has to be -2.
So the technique of partial fractions is to first deconstruct the fraction and figure out what the terms should be. Now, once we know what the terms should be, then what do we do? Well, now basically we're in great shape, because what I see now - in fact, let me show you what we have. We just did all that arithmetic. That arithmetic was just to get us to the point where we can actually look at this thing. And so if you want to integrate, the original challenge was to integrate that thing, and now what I see is that integral is identical to this integral, the integral of dx. So, using this fact, I'm able to take this integral and actually break it into two smaller integrals. And each of these integrals now is completely manageable, and I'll show you how.
So how do we do that? What we do is, in the first instance in this first integral - I'll pull out that -2, of course, out in front. So I pull out the -2 in front, then I see the integral of dx. And actually, if you make a u-substitution, letting u equal the whole bottom, then what I actually see is that this is going to be the integral of with some constant stuff here. So, in fact, this is going to give me the natural log. So when you make that u-substitution, what I think you see is . And you can check that, in fact, if you don't believe me. What's the derivative of ln blop? It's . And then what's the derivative of blop? The derivative of blop is 2, which gives me that 2 right there. So that's completely fine. So that integral is really easy, so, in fact, this equals . And the similar thing can be used here, except this is even easier. This is just . So, in fact, this challenge integral that we first saw actually can be ripped apart and we can see equals this.
So this idea of partial fractions, taking a fraction, factoring the bottom and then ripping it apart and figuring out what the constants should be, is a great technique for integrating objects that look like this.
Now, let's take a look at another one together. Let's take a look at the integral of dx. Now this is another example of our rational functions, stuff with a polynomial on top and a polynomial on the bottom, but notice the bottom now can't be factored. So the question is if the bottom can't be factored, how am I going to rip this apart into little pieces? Well what I notice is that before I even rip it apart, I notice that the power on the x's on the top is much bigger than the power of the x on the bottom. So, in some sense, maybe I should divide this out. So this is the challenge that I'm going to leave us with right now, to figure out how does one actually go about dividing that all out, figuring out exactly what this looks like. Then we can maybe integrate this. Right now, this seems like a very hard function to integrate. I don't know any technique that would actually work here. Try a u-substitution, for example, let u equal the bottom and then du equals just dx. If you let u equal the top, there's no way to untangle that, so this seems like a paradox. So let's think about a way of reducing the complexity of this by reducing the degree of the top somehow. I'll see you at the next lecture. We'll find out how it closes.

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