Integrate $\log(z − 1)/z^{n+1}$ where $n$ a positive integer, anticlockwise around the unit circle.
If we consider the principal branch of $\log(z-1)$ the integral will be very difficult to evaluate, we should consider the branch $\log(z-1) = \ln|z-1|+iarg(z)$ where $arg(z) \in (0,2\pi]$. Then $\log(z-1)$ is analytic everywhere except on the line $[1,\infty)$.
If we cut a small circular region of radius $\epsilon$ at $1$ then the integral becomes $$\int_{\partial(R-R_{\epsilon})}\dfrac{\log(z − 1)}{(z-0)^{n+1}}dz = (-1)^{n-1}\dfrac{(n-1)!}{(-1)^{n}} \cdot \dfrac{2\pi i}{n!}=-\dfrac{2\pi i}{n}$$
But I am not able to show that $\int_{\partial(R_{\epsilon})}\dfrac{\log(z − 1)}{z^{n+1}}dz \to 0$ as $\epsilon \to 0$.