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Homework Help: Integral and Partial fractions

  1. May 17, 2009 #1
    1. The problem statement, all variables and given/known data

    use the method of partial fractions on [tex]\int[/tex] [tex]\frac{36}{(x-2)(x-1)^2(x+1)^2}[/tex] dx

    2. Relevant equations

    3. The attempt at a solution

    [tex]\frac{A}{(x-2)}[/tex] + [tex]\frac{B}{(x-1)}[/tex] + [tex]\frac{C}{(x-1)^2}[/tex] + [tex]\frac{D}{(x+1)}[/tex] + [tex]\frac{E}{(x+1)^2}[/tex]
    I took the 36 out of the integral. Then

    A(x-1) (x-1)^2 (x+1) (x+1)^2 + B(x-2) (x-1)^2 (x+1) (x+1)^2 + C (x-2)(x-1) (x+1) (x+1)^2 + D(x-2) (x-1) (x-1)^2 (x+1)^2 + E (x-2)(x-1) (x-1)^2 (x+1) =1
    In the first fraction (A) (all terms cancel)
    I tried to plug in x=
    1, -1, 0, 2, -2, into (A(x-1) (x-1)^2 (x+1) (x+1)^2 + B(x-2) (x-1)^2 (x+1) (x+1)^2 + C (x-2)(x-1) (x+1) (x+1)^2 + D(x-2) (x-1) (x-1)^2 (x+1)^2 + E (x-2)(x-1) (x-1)^2 (x+1) =1)
    when x=1 I ended up with B(-2) +C(-4)=1 (I know this is probably simple algebra or I'm definitely screwing up somewhere

    Any help is appreciated
  2. jcsd
  3. May 17, 2009 #2


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    Homework Helper

    Why do you have this extra factor (x+1) in front of everything? For example for A you would get:

    [tex]\frac{A}{x-2} (x-2)(x-1)^2(x+1)^2=A (x-1)^2(x+1)^2[/tex]. You have added an extra (x+1) and you're doing that for every term it seems.

    You start out with:



    Then multiply both sides with the denominator of the right hand side.


    Now set [itex]x=1 \Rightarrow C=-\frac{1}{4}[/itex], [itex]x=2 \Rightarrow A=\frac{1}{9}[/itex] etc.
    Last edited: May 17, 2009
  4. May 17, 2009 #3


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    You made the mistake over here:
    Code (Text):
    A(x-1) (x-1)^2 (x+1) (x+1)^2 + B(x-2) (x-1)^2 (x+1) (x+1)^2 + C (x-2)(x-1) (x+1) (x+1)^2 + D(x-2) (x-1) (x-1)^2 (x+1)^2 + E (x-2)(x-1) (x-1)^2 (x+1) =1
    Because you multiple the separate fractions by the common denominator, it would end up as

    A(x-1)^2 (x+1)^2 + B(x-2)(x-1)(x+1)^2 + ... + E(x-2)(x-1)^2 = 1
  5. May 17, 2009 #4
    Thank you so much for your help zcd and cyosis. Last question: I get A= (1/9) C=(-1/4)
    But to find D and B I tried plugging in zero and didn't get anywhere. What other numbers can I plug in to solve for D and E? :blushing:
  6. May 17, 2009 #5


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    Homework Helper

    You will have to pick two values such that B and D aren't multiplied by zero. Since you know all the other constants you will get two equations relating B to D. Solve them and you're good to go.
    Last edited: May 17, 2009
  7. May 17, 2009 #6
    omg...thank you for your help...
    I did try plugging in (-2) and I end up with:
    and B(12)+D(36)=-4
    so I'm still left with the two unknowns...
  8. May 17, 2009 #7


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    Try x=0 this will give you 1/9+2B+1/2-2D+1/6=1, -> B-D=1/9. You now have two equations and two unknowns so it is solvable.
  9. May 17, 2009 #8
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