How to calculate $\oint\frac{dz}{z^3(z+4)}$ for $|z-2|<3$? Which is right $$\oint\frac{dz}{z^3(z+4)}=2\pi i(\text{Res}(f,0)+\text{Res}(f,-4))$$ or $$\oint\frac{dz}{z^3(z+4)}=2\pi i\,\text{Res}(f,0)$$? I am unsure because $z=-4$ is outside $|z-2|<3$ so does its Residue included in the value of the integral?

The second is right !!!. Namely, $\displaystyle 2\pi\,\mathrm{i}\,{1 \over 2!}\,\lim_{z \to 0}{\mathrm{d}^{2} \over \mathrm{d}z^{2}}\left[z^{3}\,{1 \over z^{3}\left(z + 4\right)}\right] = \pi\,\mathrm{i}\,\lim_{z \to 0}{2 \over \left(z + 4\right)^{3}} = \bbox[5px,border:1px dotted navy]{\displaystyle{\pi \over 32}\,\mathrm{i}}$