Complex Analysis. How to use cauchy intergral Given the following integral of gamma on the path $[0,2]$ and $[-2,2]$ we have the integral $$ \int_\gamma \frac{z} {(z^2-1)(z-3)}dz$$
I set it up like 
$$ \int_\gamma \frac{\frac{z}{(z+1)(z-3)}}{z-1}=2\pi i $$
and i get that = 
$$ \frac{-\pi i}{2} $$
and then i did the following for $[-2,2]$ but i dont believe they are set up correctly and was wondering if i can have some guidance or the error pointed out
$$ \int_\gamma \frac{\frac{z}{(z^2-1)}}{z-3}=2\pi i $$
then after everything i get its = 0 or pi i/4

If I understand your notation, by $[-2,2]$ you mean the circle of radius $2$ centered at $-2$.  But the integrand's only pole is then at $z=-1$.
So you get $\oint_Cf(z)/(z+1)$, where $f(z)=z/((z-1)(z-3))$. So the integral is equal to $2\pi if(-1)=2\pi i(-1/8)=-\pi i/4$.
