How to solve the integral $\int\frac{x-1}{\sqrt{ x^2-2x}}dx $ How to calculate  $$\int\frac{x-1}{\sqrt{ x^2-2x}}dx $$
I have no idea how to calculate it. Please help.

Substitute $u=x^2-2x,du=2(x-1)dx$. Then:
\begin{align*}
\int \frac{1}{2\sqrt{u}}du &=\frac{1}{2}\int u^{-0.5}du
\end{align*}
User the power rule:
$$\int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$$
$$\Rightarrow \frac{1}{2}\int u^{-0.5}du =\frac{1}{2}\frac{u^{-0.5+1}}{-0.5+1}$$
Back substitute and simplify:
$$\Rightarrow \frac{1}{2}\int u^{-0.5}du =\frac{1}{2}\frac{u^{-0.5+1}}{-0.5+1}=\frac{1}{2}\frac{\left(x^2-2x\right)^{-0.5+1}}{-0.5+1}=\frac{1}{2}\cdot 2\cdot\sqrt{x^2-2x}=\sqrt{x^2-2x},$$
and don't forget to add a constant $c\in\mathbb{R}$.
