How to Integrate Functions With Partial Fractions

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  • 0:11 Partial Fractions
  • 3:06 U Substitution
  • 6:32 U Substitution Example 2
  • 8:22 Lesson Summary
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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!

Partial Fractions

For the fractions to be equal, A + B must equal -1 and A - 4B must be 14
Partial Fraction Example 1

All right, let's say you want to find the area under the curve, (14-x)/(x^2 - 3x - 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 - 3x - 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 - 4B, 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 - 4B, so in order for these two fractions to be equal, A + B has to be -1, and A - 4B 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?

In the first u substitution example, the 1st term becomes 2(ln(6) - ln(2))
U Substitution 1st Term

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.

U Substitution

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 u=x - 4, and I'm going to substitute in x - 4 for u, so I'm going to integrate 1/u. I need to get rid of this dx, so I'm going to take the derivative of u=x - 4, and I get the derivative of u equals the derivative of x (du=dx). So if I plug that in, I find that my integral becomes (1/u)du. I know what the integral of (1/u)du is - it's just the natural log (ln) of u. So that is, if I take the derivative of the natural log of u, I get 1/u. But I don't want to have u in my result, I want to have x. Let's plug in x - 4 for u, so that my anti-derivative becomes ln(x - 4). If I plug that into my definite integral, I get 2ln(x - 4), evaluated from 6 to 10. Remember that when we have f(x) from 6 to 10, with this vertical bar, what we really mean is f at 10, the top of this bar, minus f evaluated at 6, the bottom of this bar. So 2ln(x - 4) from 6 to 10 is equal to 2 times the natural log of 10 - 4, or 6, minus the natural log evaluated at x=6, so that's the natural log of 2 (2ln(6) - ln(2)). This first term - remember that I have a big equation here, and this is just the first term of that equation - becomes 2(ln(6) - ln(2)).

In the first u substitution problem, the area under the curve is roughly .084
U Substitution Solution Graph

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