Evaluate: $$\int_{|z|=2} \frac{z^2}{(z-1)^2(z+1)^3} \ dz$$
The residue theorem looks promising, but the two poles are of order 2 and 3. Is there a way to compute this integrand's residues about $z = 1$ and $z= -1$ without computing repeated derivatives of $ \frac{z^2}{(z+1)^3}$, and $ \frac{z^2}{(z-1)^2}$?
Or we could just compute the Laurent series, but is there an easy way to do that? Even using partial fraction decomposition and then expanding appropriate terms as Taylor series seems cumbersome, partially because of the $z^2$ in the numerator. Is there a better way?
You may as well integrate over $|z| = R$ for any $R > 2$ (why?). What happens when $R \to \infty$? (Estimate the integrand on a large circle.)