I need to evaluate the following integral$$\int_C{\frac{z}{z^2+4z+3} dz}$$ where $C$ is the circle of centre $-1$ and radius $2$.
I found that $\frac{1}{z^2+4z+3}=-\frac{1}{2}(\frac{1}{z+3}-\frac{1}{z+1})$, $z=-3$ and $z=-1$ belongs to the circle and $f(z)=z$ is analytic on the complex plan. This leads me to
$$ \int_C \frac{z}{z^2+4z+3} dz =-\frac{1}{2}\left(\int_C{\frac{1}{z+3} dz} -\int_C{\frac{1}{z+1} dz}\right) =2\pi i $$