Exercise Evaluate $\int \frac{z^2+1}{z(z^2+4)} \, dz$ on the regions $|z|<2$ and $|z|>2$
I know that $\int_C f(z) \, dz = 2\pi i$ Res$_{z=0} \Big[ \frac{1}{z^2} f\Big( \frac{1}{z} \Big) \Big]$
So maybe we consider
$$2\pi i \, \cdot \, \, Res_{z=0} \frac{\frac{1}{z^2}+1}{z^3(z^2+4)}?$$
How can I go from here? It's not obvious to me how to find the residue of this function. Thanks in advance.
Hints:
$$\frac{\frac1{z^2}+1}{z^3(z^2+4)}=\frac{z^2+1}{4z^5}\cdot\frac1{1+\frac{z^2}4}=\frac{z^2+1}{4z^5}\left(1-\frac{z^2}4+\frac{z^4}{16}-\ldots\right)$$
Now, you only need the coefficient $\;c_{-1}\;$ of $\;z^{-1}\;$ in the above Laurent series ...