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Math 104: Calculus16 chapters | 135 lessons | 11 flashcard sets

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Lesson Transcript

Instructor:
*Sarah Wright*

In this lesson, learn how to integrate complicated fractions by using the partial fractions technique. That is, you will turn a complicated fraction into something a bit easier to integrate by finding partial fractions!

All right, let's say you want to find the area under the curve, (14-*x*)/(*x*^2 - 3*x* - 4)*dx*, and you want to find the area between *x*=6 and *x*=10. That graph looks something like this. We've got a couple asymptotes here, because at some point, this is not defined. So how do we find this area under the curve? How do we find, in particular, this integral? How do we integrate this?

Well, we want to use **partial fractions**. We've got some really nasty fraction, here, and we want to make it a little bit simpler by splitting it up. First, I'm going to factorize this quadratic, *x*^2 - 3*x* - 4, and I find that I can factor that as (*x* - 4)(*x* + 1). So I'm going to write my big fraction as being the sum of (*A*/(*x* - 4)) + (*B*/(*x* + 1)). Now *A* and *B* are undetermined coefficients. To find them, I'm going to combine these two terms, so I get *Ax* + *A* + *Bx* - 4*B*, and I did that multiplying this first term by (*x* + 1)/(*x* + 1) and the second term by (*x* - 4)/(*x* - 4). If I collect all of the terms with *x* in them, I get *x*(*A*+*B*), and the remaining terms are *A* - 4*B*, so in order for these two fractions to be equal, *A* + *B* has to be -1, and *A* - 4*B* has to equal 14. I can solve those two equations, and I get *A*=2, *B*=-3. So I can write my big, nasty fraction as being the sum of two simple fractions, that is, (2/(*x* - 4)) + (-3/(*x* + 1)). Can I integrate this? Is this going to be easier to solve?

Well, yes, because I can actually separate this into two different integrals. Remember linear properties of integrals. I can separate this into two, and pull out these constants in front of the integral. So I get 2 times the integral from 6 to 10 of (1/(*x* - 4))*dx* - 3 times the integral from 6 to 10 of (1/(*x* + 1))*dx*. Okay, can I find these integrals? Well, let's look at that first term.

I have 2 times the integral from 6 to 10 of (1/(*x* - 4))*dx* - hmm. Well, let's use a ** u substitution** here. Let's say that

What about this second term, this -3 times the integral from 6 to 10 of (1/(*x* + 1))*dx*? Well, we're going to do the same thing. I'm going to do a *u* substitution, where *u*=*x* + 1. If I take the derivative of that, I get *du*=*dx*. Let me plug those into my integral, and I get the integral of (1/*u*)*du* - that's just the natural log. We see these a lot when we're dealing with partial fractions; we're almost always going to have a natural log. So ln(*u*) is equal to ln(*x* + 1) when I plug in my *x* + 1 for *u*. This entire term becomes 3ln(*x* + 1) evaluated from 6 to 10. That equals 3(ln(11) - ln(7)). If I plug those in for my second term, I find that my entire integral is equal to 2(ln(6) - ln(2)) - 3(ln(11) - ln(7)). That's roughly .84 if I plug it into a calculator. So in this case, the area under the curve is roughly equal to .084.

So let's do another example. Let's say I want to integrate (2*x*)/(*x*^2 - 9) from *x*=4 to *x*=5. I could factor *x*^2 - 9, and get (*x* - 3)(*x* + 3). But wait a second. I'm a big fan of partial fractions, but you should really only use partial fractions when an easier method won't work. And if I take a really good look at this, I see that I can actually do a *u* substitution on this fraction. Humor me here.

Let's say that I'm going to substitute *u* for *x*^2 - 9. That means that *du* would be the derivative of this, so that's 2*xdx*. That means that I've got (1/*u*)*du*, and the integral of that is just ln(*u*). If I plug in *x*^2 - 9 for *u*, I find that my integral is ln(*x*^2 - 9). Wow! That was a lot easier than doing the partial fraction! Let's use this. So ln(*x*^2 - 9), evaluated from *x*=4 to *x*=5, gives me the natural log of (5^2) - 9, which is ln(16), minus the natural log of (4^2) - 9, or ln(7). Well this equals .83. So solving it by *u* substitution was a lot faster than going through and finding the partial fraction, although you could have done that.

So, when you're **integrating complex fractions**, the first thing you want to do is see if a substitution will work - it'll save you a lot of time in the long run. If it doesn't work, you want to factor the fraction and use undetermined coefficients to break up this complex fraction into some easier, smaller fractions. On each one of these, you're probably going to need *u* substitution, but then you can usually integrate very easily. Just remember, with complex fractions (and, really, all integration problems), you might have to try a few things. See if *u* substitution works. If it doesn't, use partial fractions. If, say, partial fractions don't work, maybe you need to integrate by parts. It's all going to be a little trial and error, but you'll get the hang of it.

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Math 104: Calculus16 chapters | 135 lessons | 11 flashcard sets

- Go to Continuity

- Go to Series

- Go to Limits

- Calculating Integrals of Simple Shapes 7:50
- Anti-Derivatives: Calculating Indefinite Integrals of Polynomials 11:55
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- How to Solve Integrals Using Substitution 10:52
- Substitution Techniques for Difficult Integrals 10:59
- Using Integration By Parts 12:24
- Partial Fractions: How to Factorize Fractions with Quadratic Denominators 12:37
- How to Integrate Functions With Partial Fractions 9:11
- How to Use Trigonometric Substitution to Solve Integrals 13:28
- How to Solve Improper Integrals 11:01
- Go to Integration and Integration Techniques

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