How to calculate this contour integral? 
Calculate $$\oint_C \frac{dz}{(z^2+9)(z+9)}$$ with $C: |z|=4$

I know that the function $\frac{1}{(z^2+9)(z+9)}$ is analytic in $\mathbb{C}$ except in the points $3i,-3i,-9$.
I've tried use the Cauchy's Formula but i don't find the way to use it.
I appreciate your collaboration.
 A: Here's a more scenic route
$$\oint_{|z|=4}\frac{1}{(z^2+9)(z+9)}\ \mathrm dz$$
After some partial fraction decomposition, followed by an application of the estimation lemma, we have
$$\frac{1}{90}\oint_{|z|=4}\frac{9-z}{z^2+9}\ \mathrm dz$$
We can further decompose this fraction to again find another contour integral that vanishes. Which leaves us with
$$-\frac{1}{180}\oint_{|z|=4}\frac{2z}{z^2+9}\ \mathrm dz$$
$$=-\frac{1}{180}\oint_{\gamma}\frac{1}{w}\ \mathrm dw$$
Where
$$\gamma=16e^{2it}+9$$
$$0\leq t\leq2\pi$$
Note that the contour $\gamma$ winds around the origin twice in a counterclockwise direction, which implies that
$$-\frac{1}{180}\oint_{\gamma}\frac{1}{w}\ \mathrm dw=-\frac{4\pi i}{180}=-\frac{\pi i}{45}$$
Therefore
$$\oint_{|z|=4}\frac{1}{(z^2+9)(z+9)}\ \mathrm dz=-\frac{\pi i}{45}$$
A: HINT
Given $$\int_c \frac{dz}{(z-3i)(z+3i)(z+9)}$$
The counter is $|z|=4$ so the sigularities in counter are $3i$ and $-3i$ apply cauchy residue theorem
$$\int_c \frac{dz}{(z-3i)(z+3i)(z+9)}=2\pi i \left \{ Res(f,3i)+Res(f,-3i)\right \}$$
A: Cauchy's residue theorem is the way to go, but if you haven't seen that in your course yet, you can still use Cauchy's integral formula.
Your contour $C$ contains the singularities $\pm3i$, but for CIF you can only
deal with one at a time. The trick is to "split" the circular
contour $C$ into two smaller contours, each containing one singularity.
So let $C_+$ and $C_-$ denote the semicircles, in the upper and lower half
plane respectively, centred at $0$ with radius $4$. Then
$$\int_C f(z)\,dz=\int_{C_+} f(z)\,dz+\int_{C_-} f(z)\,dz$$
as the integrals over the straight-line segments cancel. Each of these
new integrals can be done with CIF:
$$\int_{C_+}\frac{dz}{(z^2+9)(z+9)}=\int_{C_+}\frac{g(z)\,dz}{z-3i}
=2\pi i\,g(3i)$$
where
$$g(z)=\frac{1}{(z+3i)(z+9)}$$
etc.
