Cauchy's integral formula is:
$$f(z_0)n(\gamma,z_0)=\frac{1}{2\pi i}\oint_\gamma\frac{f(z)}{z-z_0}dz$$
The problem I had to solve was:
$$\int_\gamma \frac{dz}{(z^2+9)(z+9)};\gamma:|z|=4$$
My thoughts were that we could write: $f(z)=\frac{1}{z^2+9}$ and $z_0=-9$, therefore $z_0$ is outside of $\gamma$ and as such $n(\gamma,-9)=0$. Therefore:
$$\int_\gamma \frac{dz}{(z^2+9)(z+9)}=0$$
Since whenever a result in math is $0$ you have to be wary, I'm asking here.
That integral is equal to $-\frac{\pi i}{45}$. In fact, the function $f$ has $2$ and only two singularities at the region bounded by the curve $|z|=4$: at $\pm3i$. Since$$\operatorname{res}_{z=3i}f(z)=-\frac1{180}-\frac i{60}\quad\text{and}\quad\operatorname{res}_{z=-3i}f(z)=-\frac1{180}+\frac i{60},$$your integral is equal to$$2\pi i\left(-\frac1{180}-\frac i{60}-\frac1{180}+\frac i{60}\right)=-\frac{\pi i}{45}.$$The problem with your approach is that $f$ is not defined at all points of the region bounded by the curve $|z|=4$.
You can also solve the problem doing a partial fraction decomposition. It turns out that$$f(z)=\frac\alpha{z+9}+\frac\beta{z-3i}+\frac\gamma{z+3i}$$with$$\alpha=\frac{1}{90},\quad\beta=-\frac{1}{180}-\frac{i}{60},\quad\text{and}\quad\gamma=-\frac{1}{180}+\frac{i}{60}.$$Now, you can apply Cauchy's integral formula.