Contour Integral $\frac{z}{z^2+4z+3}$ where C is the circle with center -1 and radius 2 counterlockwise

Question

I'm trying to determine the contour integral of the equation above. I first split the fraction into $g(z)=\frac{z}{z+3}\frac{1}{z+1}$

Then, since $z_0=-1$ is enclosed in the circle $|z+1|=2$, where -1 isn't analytic, I want to focus on $f(z)=\frac{z}{z+3}$.

Next, $\oint \frac{z}{z+3}dz=2\pi if(-1)=2\pi i\frac{-1}{-1+3}=-\pi i$

Are these steps correct? Is there a way to verify this?

Answer 1

HINT:Integral $I=2πi×Res[f(z),z=-1]+πi× Res[f(z),z=-3]$

Edit-A more general form of Cauchy's residue theorem:

$\int_C f(z)dz=2πi.\sum Res[f(z);z=z_k]+πi.\sum Res[f(z);z=z_c]$

where $z_k$ is a pole inside contour $C$ and $z_c$ is a pole on the boundary of contour $C$.