How to evaluate the integral $\oint_{C}\frac{z + 2}{z^2(z^2-1)}dz$ I've been trying to solve the following complex-analysis problem:
Evaluate 
$\oint_{C}\frac{z + 2}{z^2(z^2-1)}dz$
Where $C = \{z : |z + 1| = \frac{3}{2}\}$
What I've tried to do was the following: I detected that there are singularities at $z = -1$ and at $z = 0$, making the function non-analytic inside that domain. So I found the Laurent Series for $g(z) = z + 2$ around $-1$, which is $g(z) = 1 + (z - (-1))$, but from that point I  can't find a way to use the coefficients $a_n$ to solve the integral with the Residue Method. I was not able to transform the function $g(z)$ into some $\frac{h(z)}{(z - (-1))} = \frac{z + 2}{z^2(z^2-1)}$ with $a_n$'s that are not functions of $z$.
 A: Let us denote
$$f(z)=\frac{z+2}{z^2(z^2-1)}.$$
Then for the Laurent expansion at $z=-1$ we rewrite this function as
$$f(z)=\frac{1}{z+1}\cdot\underbrace{\frac{z+2}{z^2(z-1)}}_{g(z)}=\frac{1}{z+1}\left[g(-1)+\frac{g'(-1)}{1!}(z+1)+\ldots\right].$$
Similarly, for the expansion at $z=0$ we write it as
$$f(z)=\frac{1}{z^2}\cdot\underbrace{\frac{z+2}{z^2-1}}_{h(z)}=\frac{1}{z^2}\cdot\left[h(0)+\frac{h'(0)}{1!}z+\ldots\right]$$
Now it becomes clear that the residues at $z=-1$ and $z=0$ are $g(-1)$ and $h'(0)$. I think you can calculate them from there.
A: You should expand $$\frac{z+2}{z^{2}(z^{2}-1)}$$ as $$\frac{1}{z(z^{2}-1)}+2(\frac{1}{z^{2}-1}-\frac{1}{z^{2}})=\frac{1}{2z}(\frac{1}{z-1}-\frac{1}{z+1})+(\frac{1}{z-1}-\frac{1}{z+1})-\frac{2}{z^{2}}$$where the first part can be further expanded as $$\frac{1}{2}(\frac{1}{z-1}-\frac{1}{z}-(\frac{1}{z}-\frac{1}{z+1}))=\frac{1}{2}(\frac{1}{z-1}+\frac{1}{z+1}-\frac{2}{z})$$So we are evaluating $$\frac{1.5}{z-1}-\frac{0.5}{z+1}-\frac{1}{z}-\frac{2}{z^{2}}$$
As others pointed out, the integral should be $-1.5*2\pi i=-3\pi i$. 
