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Finding Partial Fractions

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Find partial fractions for 4/(x^3-2x^2)

    3. The attempt at a solution
    Heres the steps that I took:
    1. 4/(x^3-2x^2)= 4/(x^2(x-2))= A/(x^2) + B/(x-2)
    2. 4= A(x-2) + B(x^2)
    3. When x=0, -2A=4, so A=-2,
    and When x=2, 4B=4, so B=1.
    4. So my final answer was:
    -2/(x^2)+1/(x-2)

    The real answer as I found out from Wolfram Alpha integral calculator was:
    -2/(x^2)+1/(x-2)-1/x

    So the real answer is the same as the solution that I got, except for the -1/x at the end... I have no idea where that -1/x came from, no matter how many times I redo this problem. Please tell how to get the real answer! Thank you!
     
  2. jcsd
  3. Nov 20, 2011 #2

    LCKurtz

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    I assume you know yours is incorrect because it doesn't expand to equal the given fraction.

    The correct expansion is
    [tex]\frac{4}{x^2(x-2)}=\frac{Ax+B}{x^2}+\frac{C}{x-2}[/tex]
     
  4. Nov 20, 2011 #3

    SammyS

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    You should have something like: [itex]\displaystyle \frac{4}{x^2(x-2)}=\frac{Ax+B}{x^2}+\frac{C}{x-2}\frac{}{}[/itex]

    or equivalently: [itex]\displaystyle \frac{4}{x^2(x-2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-2}\frac{}{}[/itex]
     
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