Integration using residues $\int z^2 \log [(z+1)/(z-1)] dz$

Question

$\int z^2 \log [(z+1)/(z-1)] dz$ taken around circle $|z|=2$

I am taking residues at $\pm 1$.

This gives me 0 as the value of integral. Is this correct.

How do I modify the integral to get value over half the circle?

Answer 1

$$\int_C f(z)dz=-2\pi i \operatorname{Res}_{z=\infty}f(z)$$ But we have $$f(z)=2z+\frac23z^{-1}+o(z^{-1})$$ as $z\to\infty$, therefore the integral equals $\frac43 \pi i$. $f$ is not meromorphic in $|z|<2$ , so we can not apply residue theorem inside the circle, but we can apply it outside the circle.