With the assistance of the local Cauchy Integral formula find the values of:
(i) $\displaystyle\int_{|z-2|=2}(z^2-1)^{-1}\tan^{-1}(z)dz$;
(ii)$\displaystyle\int_{|z|=2}(z+1)^{-2}e^zdz$.
I tried to solve these two integrals, the first one I do not know if I solved it well and the second one I do not know how to solve it, could someone help me please? Thank you very much, here is my attempt to solve (i): \begin{align} \int_{|z-2|=2}(z^2-1)^{-1}\tan^{-1}(z)dz &= \int_{|z-2|=2}\frac{\tan^{-1}(z)}{z^2-1}dz \\ &= \int_{|z-2|=2}\frac{\tan^{-1}(z)}{2(z-1)}-\frac{\tan^{-1}(z)}{2(z+1)}dz \\ &= \int_{|z-2|=2}\frac{\tan^{-1}(z)}{2(z-1)}dz \\ &= \frac{1}{2}2\pi i\tan^{-1}(1) \\ &= \dfrac{\pi^2i}{4} \end{align}