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.

Notice, 
$$\int \frac{x-1}{\sqrt{x^2-2x}}\ dx=\frac 12\int \frac{2(x-1)}{\sqrt{x^2-2x}}\ dx $$
$$=\frac 12\int \frac{d(x^2-2x)}{\sqrt{x^2-2x}}$$
$$=\frac 12\int (x^2-2x)^{-1/2}d(x^2-2x)$$
$$=\frac 12\frac{(x^2-2x)^{-\frac 12+1}}{-\frac 12+1}+C$$
$$=\frac 12\frac{(x^2-2x)^{1/2}}{1/2}+C$$
$$=\color{red}{\sqrt{x^2-2x}+C}$$
