1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration using partial fractions

  1. May 31, 2010 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations


    if x-r occurs with multiplicity m, then A/(x-r) must be replaced by a sum of the form:

    I think this second equation is the source of my confusion.

    3. The attempt at a solution

    I began by factoring the denominator:
    x4-x2 = x2(x+1)(x-1)

    So, according to my book, we have the following constants:
    A/x + B/x2 + C/(x+1) + D/(x-1)

    First question: where did the x come from in the constant A/x? Does this follow from the rule above? That is, because x2 has multiplicity 2, I get A/x and B/x2?

    Next, according to my book, when you clear the fractions, you should get:
    Ax(x+1)(x-1) + B(x+1)(x-1) + Cx2(x-1) + Dx2(x+1)

    I don't understand this. Why isn't it Ax2(x+1)(x-1) + Bx(x+1)(x-1) + Cx3(x-1) + Dx3(x+1)?

    Can someone explain? Many thanks.
  2. jcsd
  3. May 31, 2010 #2


    User Avatar
    Homework Helper

    you multiply everything by the original denominator to get the original fraction

    try multiplying out below and it should all become clear ;)
    [tex] \frac{(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} + \frac{D}{x-1})(x^2(x+1)(x-1))
    } [/tex]
    Last edited: May 31, 2010
  4. May 31, 2010 #3


    Staff: Mentor

    A, B, C, and D are constants, but the expressions above aren't constant.
    Yes. The denominator in factored form has an x2 factor, so the multiplicity of this factor is 2. This means that you'll need to have terms A/x and B/x2.
    You have this equation, which has to be identically true (true for all values of x other than those that make any denominator 0).

    [tex]\frac{3x^3 - 4x^2 - 3x + 2}{x^2(x^2 - 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 1} + \frac{D}{x - 1}[/tex]

    You can clear the fractions by multiplying both sides of this equation by x2(x2 - 1). When you multiply the terms on the right hand side above, there will be some cancellation.
  5. Jun 1, 2010 #4
    That clears it up. Thanks, guys.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook