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.
 A: $$\int\frac{x-1}{\sqrt{ x^2-2x}}dx=\frac12\int\frac{2x-2}{\sqrt{ x^2-2x}}dx$$
Substitute $u=x^2-2x,\;du=(2x-2)dx$ to get:
$$\frac12\int\frac{1}{\sqrt{u}}du=\frac12\int u^{-1/2}du=\sqrt{u}+C=\sqrt{x^2-2x}+C$$
A: $$t=x^2-2x$$
$$dt=2x-2dx$$
$$dx=\frac{dt}{2(x-1)}$$
$$\int\frac{x-1}{\sqrt{x^2-2x}}dx=\int\frac{1}{2}\frac{1}{t}dt=\frac{1}{2}\int t^{\frac{-1}{2}}=\frac{1}{2}2t^{\frac{1}{2}} +C = \sqrt{x^2-2x}+C$$
A: 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 $c\in\mathbb{R}$.
A: Notice, 
$$\int \frac{x-1}{\sqrt{x^2-2x}}\ dx=\frac 12\int \frac{2(x-1)}{\sqrt{x^2-2x}}\ dx $$
$$=\frac 12\int \frac{d(x^2-2x)}{\sqrt{x^2-2x}}$$
$$=\frac 12\int (x^2-2x)^{-1/2}d(x^2-2x)$$
$$=\frac 12\frac{(x^2-2x)^{-\frac 12+1}}{-\frac 12+1}+C$$
$$=\frac 12\frac{(x^2-2x)^{1/2}}{1/2}+C$$
$$=\color{red}{\sqrt{x^2-2x}+C}$$
A: let $(x^2-2x)=t^2$. Then $(x-1)dx =t dt$ in numerator replace $(x-1)dx$ by $t\ dt$. In denominator put $t$, cancel $t$ by $t$ then you will find integral $dt$ ie. equal to $t$. (ANSWER)
A: we have,
$$I=\int \frac{x-1}{\sqrt(x^2 -2x)}dx----------(1)
    $$
put ,              $$ (x^2 -2x)=t $$
                      on differentiating, we get,    $$ 2xdx-2dx=dt$$
$$2dx(x-1)=dt$$
$$dx(x-1)=\frac{dt}{2}$$
substitute on equation (1) we get,
$$I=\int \frac{\frac{1}{2}dt}{\sqrt(t)} $$
$$I=1/2 \int t^-(\frac{1}{2}) dt $$
$$I=1/2 \frac{t^(-\frac{1}{2} +1)}{(-\frac{1}{2} +1)} + C $$
$$I=\frac{1}{2} \frac{t^\frac{1}{2}}{\frac{1}{2}} + C$$
now substitute the value of t,
$$I=\sqrt(x^2-2x) + C-------(2)$$
equation (2) is the required integration of equation (1)
