calculate $\int_C \frac{z^4+2z}{(z-i)^3}dz$ where $i$ is interior to $C$.

Question

So I need to find

$$\int_C \frac{z^4+2z}{(z-i)^3}dz.$$

Here, $i$ is inside of $C$.

Attempt:

By Cauchy's integral formula, we have

\begin{align} \int_C \frac{z^4+2z}{(z-i)^3}dz&=\frac{2\pi i}{2!}\bigg(f^{''}(z^4+2z) \bigg)\bigg\vert_{z=i}\\ &=\pi i (12z^2)\bigg\vert_{z=i}\\ &=-12\pi i. \end{align}

Am I correct?