$\int_{\alpha}z^2 \log\left(\frac{z+1}{z-1}\right)\,dz$ where $\alpha$ is $|z-1|=1$ Calculate
$$\int_{\alpha}z^2 \log\left(\frac{z+1}{z-1}\right)\,dz$$
where $\alpha$ is $|z-1|=1$ and the initial point of integration $z_1=1+i$.

The function isn't analytic at $1$ and $-1$. Is there a theorem to solve it?
Transformation of the integral $z\to 1/z$
$$f(z)=\frac{\log\left(\frac{1+z}{1-z}\right)}{z^4}$$
How do I transform the circumference $1+e^{i\theta}$?
Is there any other way?

$g(z)=z^2 (\log(1+z)-i\pi)$ is analytic on $|z-1| < 2$ so that $\int_{|z-1| = 1} g(z)dz = 0$. 
Thus, it reduces to finding a primitive of $$h(z) = -z^2 \log(1-z) = \sum_{n=1}^\infty \frac{z^{n+2}}{n}$$ ie. $$H(z) = \sum_{n=1}^\infty \frac{z^{n+3}}{n(n+3)}=\frac13\sum_{n=1}^\infty z^{n+3} (\frac{1}{n+3}-\frac{1}{n}) \\ =-\frac13 z^3\log(1-z)+ \frac13 (\log(1-z)-z-\frac{z^2}{2})$$ 
Finally, being careful with the branches, and using that $H(1+ze^{2i\pi}) =H(1+z)- \frac{z^3-1}{3}2i\pi$
 $$\int_{|z-1|=1} z^2 \log\left(\frac{z+1}{z-1}\right)\,dz=\int_{|z-1|=1} (g(z)+h(z))dz = \int_{|z-1|=1} h(z)dz  = H(1+1)-H(1+ e^{2i\pi})\\= \frac{7}{3} 2i \pi$$
or something like that
