How to Find the Partial Fraction of z/(z^2 -1)?

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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.
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Assume you have a, b where

\frac{z}{z^2-1}=\frac{a}{z-1}+\frac{b}{z+1}

Solve for a and b anyway you like.
 
Many thanks!
 
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