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That Might Be Improper
Integration of Partial Fractions

 

 When we want to find an integral like this one ...

 

 
 we have a big problem.
 Up to now, our only trick with stuff like this was the old ...
 

 

 
 routine.
 The odds of that working on something this big are not so good.
 In this example, we have a bigger exponent in the numerator
 than anything in the denominator,
 so there's NO chance of it working.
 
 A fraction like this one, with a larger power of X in the numerator
 than any we have in the denominator,
 is called an improper fraction.
 
 If the biggest exponent in a fraction is in the denominator,
 it is called a proper fraction.
 They're easier to deal with, calculus wise anyway ...
 
 WELL WE DON'T HAVE A PROPER FRACTION,
 WHAT DO WE DO NOW?
 
 Just turn it into a proper fraction of course!
 
 Oh yeah? How do we do that?
 
 We use our good buddy, long division.
 And with what we've got here, we're talkin' LONG division.
 

 
 So that means ...
 

 
 Most of this new thing is easy to deal with.
 We'll go as far as we can.
 We get ...
 

 
 We still have a tough part, but now it's a lot smaller.
 We still can't use the lnX trick (we got close!).
 We don't have a "+ C" part yet,
 we can say it's still hiding inside the last integral part.
 
 Now we deal with the last part by doing a strange thing.
 We factor the denominator!
 

 
 What we want to do is turn this last term into two fractions.
 Each fraction will have a constant in the numerator,
 and one of our two factors in the denominator.
 
 We want to have ...
 

 
 Once upon a time (hopefully) you learned how to add two fractions
 by finding a common denominator.
 Here we are doing just the opposite.
 We're ripping a single fraction apart into two pieces that,
 we hope, will be easier to deal with.
 
 The next step to solve this is to clear the denominators.
 To do that, multiply both sides by (X - 1)(X - 2).
 You get ...
 

 2X + 12 = A(X - 2) + B(X - 1)

 
 Multiply out the right side ...
 

 2X + 12 = AX - 2A + BX - B

 
 Rearrange a bit ...
 

 2X + 12 = AX + BX - 2A - B

 
 Now factor the right side ...
 

2X + 12 = (A + B)X - 2A - B

 
 Now here's a great trick.
 Since the left side of the equation is equal to the right.
 (Hey, that's what an equation does!)
 That means the coefficient on the X term on the left side (the 2),
 is equal to the coefficient on the X term on the right side (the A + B)!
 So ...

2 = A + B 

 
 and, and, and ...
 By the same logic, the number on the left (12)
 MUST be equal to the numbers on the right!
 So ...

 12 = -2A - B

 
 So now we have two equations with two unknowns.
 That means we can solve for them ...
 

 
 We can solve for B by substitution ...
 

 
 So A = -14, and B = 16.
 That means our remaining integral part (remember the problem?)
 is ...
 

 

 
 so cleaning up these two a bit ...
 

 
 And now we finally get to use the "ln" thing!
 

 
 Remember, something like ln-5 doesn't mean anything,
 so we use absolute value notation to make sure we get
 ln(something positive).
 Tacking this on to the part we already solved, we have ...
 

 
 We might have expected this one to have a long and messy answer!
 
 Review Time:
 
 When we have an improper fraction,
 we divide the numerator by the denominator
 to get some terms all by themselves 
 AND a proper fraction.
 
 And now we also have a way to integrate a proper fraction
 even when it isn't a du/u setup to begin with.
 

 
 There is another way that you can use to figure out A and B.
 
 We'll use the proper fraction part from the last example
 so that you see we get the same answers.
 

 
 First, factor the denominator of the fraction ...
 

 
 This time, we cover up one of the factors of the denominator ...
 

 
 The X value that makes the covered up factor equal to zero is 1.
 So we solve the rest of the fraction for X = 1 ...
 

 
 This is the numerator of one of our integrals.
 The factor that we covered up, is the denominator.
 So that part of the integral is ...
 

 
 To find the other piece, cover the other denominator factor
 and solve the rest for the X value that makes THAT factor
 equal to zero (X = 2) ...
 

 
 So the other integral term is ...
 

 
 You can check these with what we got before if you want.
 It's the same integrals.
 
 Let's do another one ...
 
 Example:
 

 
 The first thing that we need to do, 
 is check to see if this is a proper fraction or not.
 The numerator is degree 2 (highest exponent term is 3X 2).
 The denominator is degree 3 (highest exponent term is X 3).
 That means we have a proper fraction.
 We get a big break, because the denominator has already been factored. 
 
 Simplify this thing using the original method ...
 

 
 Multiply both sides by (X - 1)(X - 2)(X - 3).
 We get ...
 

 3X2 - 5X + 2 = A(X-2)(X-3) + B(X-1)(X-3) + C(X-1)(X-2)

 
 Multiply it all out ...
 

 3X2 - 5X + 2 = AX2 - 5AX + 6A + BX2 - 4BX + 4B + CX2- 3CX + 2C

 
 Rearrange the right side ...
 

 3X2 - 5X + 2 = (A + B + C)X2 + (- 5A - 4B - 3C)X + (6A + 3B + 2C)

 
 So from the coefficients we have ...
 

A + B + C = 3

 

 - 5A - 4B - 3C = - 5

 

6A + 3B + 2C = 2

 
 Solve these any way you like, you get ...
 

A = 0    B = - 4    C = 7

 
 So if we can remember which letter 
 was over which part of the denominator,
 we have ...
 

 
 Using the "cover up" method, we would do this ...
 
 Step 1, find A ...
 

 
 Step 2, find B ...
 

 
 Step 3, find C ...
 

 
 These are the same answers we got using the other method.
 Most people find this method easier and faster to use,
 but use either one you like.
 Neither one is any more correct than the other.
 They are both just algebra tricks.
 

   copyright 2005 Bruce Kirkpatrick

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