How do I integrate $\frac1{(x-1)^3}$ and prove it equals $-\frac1{2(x-1)^2}$

Question

In calculus, how do I proof the integral of $\dfrac{1}{(x-1)^3}$ is $-\dfrac{1}{2(x-1)^2}$.

I know how I can derive $\dfrac{1}{2(x-1)^2}$ to $-\dfrac{1}{(x-1)^3}$, but I can't seem to do integrate the other way.

Thank you

Answer 1

$$\frac{1}{(x-1)^3}=(x-1)^{-3}=1\cdot (x-1)^{-3}=\underbrace{x'\cdot(x-1)^{-3}=\left[\frac{(x-1)^{-2}}{-2}\right]'}_{\displaystyle \int u'(x)(u(x))^k\,dx=\frac{(u(x))^{k+1}}{k+1}\text{, for any }k\neq -1}$$