integrating $\oint_C \frac{3z^3 + 2}{(z-1)(z^2 + 9)}dz$ on $|z|=4$ I am doing $\oint_C  \dfrac{3z^3 + 2}{(z-1)(z^2 + 9)}dz$ on $|z|=4$ and I find that there are poles within the contour at $z = 1$ and at $z = 3i$, both simple poles. I find that the integral $I = 2\pi i  \,\text{Res}(1) + 2 \pi i \,\text{Res}(3i)$, or $I = \pi i + 2 \pi i \,\,\,\text{Res}(3i)$, with only that second residue left to find. Is this correct, or should I include  $\text{Res}(-3i)$ as well?

Let $\,C_1,\,C_2,\,C_3\,$ be little circles (say, of radius $\,0.1\,$) around each of the poles $\,1,\,-3i,\,3i\,$ resp., of the function, so by Cauchy's Integral Theorem:
$$\int\limits_{|z|=4}\frac{3z^3+2}{(z-1)(z^2+9)}dz=\int\limits_{C_1}\frac{\frac{3z^3+2}{(z^2+9)}}{z-1}dz+\int\limits_{C_2}\frac{\frac{3z^3+2}{(z-1)(z-3i)}}{z+3i}dz+\int\limits_{C_3}\frac{\frac{3z^3+2}{(z-1)(z+3i)}}{z-3i}dz=$$
$$=2\pi i\left(\left.\frac{3z^3+2}{(z^2+9)}\right|_{z=1}+\left.\frac{3z^3+2}{(z-1)(z-3i)}\right|_{z=-3i}+\left.\frac{3z^3+2}{(z-1)(z+3i)}\right|_{z=3i}\right)=$$
$$=2\pi i\left(\frac{1}{2}+\frac{2+81i}{-18+6i}+\frac{2-81i}{-18-6i}\right)=$$
$$=2\pi i\left(\frac{1}{2}-\frac{1}{6}\left(-5\right)\right)=\frac{8\pi i}{3}$$
