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Partial fraction of integrand

  1. Nov 25, 2004 #1
    As part of finding the integral of z/(z^2 -1), I'm stuck on getting the partial fraction for it. 1/2 [(1/(z-1) - 1/(z+1)] gives 1/(z^2-1). What should I do to get the z in the numerator. Any hints welcome.
    Regards
     
  2. jcsd
  3. Nov 25, 2004 #2

    shmoe

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    Assume you have a, b where

    [tex]\frac{z}{z^2-1}=\frac{a}{z-1}+\frac{b}{z+1}[/tex]

    Solve for a and b anyway you like.
     
  4. Nov 25, 2004 #3
    Many thanks!
     
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