The following two facts are quite useful for computing residues:

If $f$ has a pole of order at most $n+1$ at $x$ , then $$ \operatorname{Res}(f;x) = \lim\limits_{y \to x} {1 \over n!} {d^n \over dy^n} \left( (y-x)^{n+1} f(y) \right) $$

If $g$ is regular at $x$ and $f$ has a simple pole at $x$ , then $\operatorname{Res}(fg;x) = g(x) \operatorname{Res}(f;x)$ .