Given the following integral of gamma on the path for the circle $[0,2]$ and circle $[-2,2]$. Where 0 is the center and 2 is the radius as well for the 2nd circle 0,-2 is the center and 2 is the radius. 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