# Partial fraction of integrand

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

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