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But Wait! There's More ...
Integration of More Partial Fractions

 

 When we need to find the integral of a fraction

 where one of the factors in the denominator is repeated,
 there is a complication.
 

 
 The denominators we wind up with might be a surprise. 
 They are (X + 3), (X - 2) and ...
 (X - 2) 2.
 

 
 Before we go on, there is some bad news.
 The "cover up the factor" trick won't work
 when two or more of the denominator factors are the same.
 That means we are left with the standard method.
 Step one of the standard method is to get rid of the denominator.
 To do it, multiply both sides by (X - 2) 2(X + 3).
 We get ...
 

5X 2 - 9X - 22 = (X - 2)(X + 3)A + (X + 3)B + (X - 2) 2C

 
 Multiply out the right side ...
 

5X 2 - 9X - 22 = AX 2 + AX - 6A + BX + 3B + CX 2 - 4CX + 4C

 
 Gathering and factoring ...
 

5X 2 - 9X - 22 = (A + C)X 2 + (A + B - 4C)X + (- 6A + 3B + 4C)

 
 So ...
 
A + C = 5
 
A + B - 4C = - 9
 
- 6A + 3B + 4C = - 22
 
 Solve these any way you like,
 you get ...
 

A = 3      B = - 4      C = 2

 
 So we have ...
 

 
 Part 1:

 
 Part 2:

 
 Part 3:

 
 Putting it all together, we get ...
 

 

 
 You might have a fraction 
 where the denominator can't be factored out 
 to linear terms (that means no exponents on the X's).
 We might have ...
 

 
 The first thing to see is that the degree of the numerator (2),
 is less than the degree of the denominator (3).
 That means that this is a proper fraction.
 If it wasn't, we would multiply out the denominator
 and do long division on this puppy.
 Since this is a proper fraction already, long division won't help.
 We start in with the A & B stuff.
 This time, because we have that non linear term in the denominator,
 the terms break apart like this ...
 

 
 Obviously, the big deal here is the AX + B in the numerator.
 In the A & B terms we have had before this,
 the exponent in the denominator has always been 1.
 We did not have any X's in the numerators.
 Another way to say that, is that the exponent on the X in the numerator
 was ONE LESS in the numerator than in the denominator.
 That is, X to the zero power.
 What we have here then is the same deal.
 The exponent on the X in the numerator is one less 
 than the exponent on the X in the denominator.
 
 So much for the explanation, on with the problem!
 
 The first step is to multiply both sides to get rid of the denominator.
 That means we multiply both sides by (X 2 + 1)(X - 3).
 

3X 2 - 7X - 16 = (AX + B)(X + 3) + (X 2 + 1)(C)

 
 Multiply out the right side ...
 

3X 2 - 7X - 16 = AX 2 - 3AX + BX - 3B + CX 2 + C

 
 Rearrange and refactor ...
 

3X 2 - 7X - 16 = (A + C)X 2 + (- 3A + B)X +(- 3B + C)

 
 So we have ...
 
A + C = 3
 
- 3A + B = - 7
 
- 3B + C = - 16
 
 Solving these we get ...
 

 A = 4      B = 5      C = -1

 
 That makes our integrals ...
 

 
 Let's look at the first integral.
 Split it into two pieces ...
 

 
 The first part is easy ...
 

 
 Since X 2 + 1 can't be negative, we can drop the absolute value lines.
 

 
 The second part looks like a trig substitution problem.
 Remember them?
 

 

 

 
 and from tan q = X we get q = tan -1X, 
 so ...

 

 
 That's two thirds of the problem done ...
 

 
 For the last part, we substitute: u = X - 3, du = dX:
 

 
 And we're done!
 
 Wow! What a trip! The original integral was ...
 

 
 and our answer is ...
 

 
 Don't get too impressed with all this substitution
 and reduction formula stuff we've been doing 
 for the last few pages.
 There are lots of integrals that can't be solved
 with any of these tricks.
 Today, we can grind some of them through computers.
 
 Long before there were computers, though,
 there were Brook Taylor and Colin McLaurin ...
 

   copyright 2005 Bruce Kirkpatrick

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