How to solve the integral $\int\frac{x-1}{\sqrt{ x^2-2x}}dx $

Question

How to calculate $$\int\frac{x-1}{\sqrt{ x^2-2x}}dx $$

I have no idea how to calculate it. Please help.

Answer 1

$$t=x^2-2x$$ $$dt=2x-2dx$$ $$dx=\frac{dt}{2(x-1)}$$ $$\int\frac{x-1}{\sqrt{x^2-2x}}dx=\int\frac{1}{2}\frac{1}{t}dt=\frac{1}{2}\int t^{\frac{-1}{2}}=\frac{1}{2}2t^{\frac{1}{2}} +C = \sqrt{x^2-2x}+C$$

Answer 2

$$\int\frac{x-1}{\sqrt{ x^2-2x}}dx=\frac12\int\frac{2x-2}{\sqrt{ x^2-2x}}dx$$

Substitute $u=x^2-2x,\;du=(2x-2)dx$ to get:

$$\frac12\int\frac{1}{\sqrt{u}}du=\frac12\int u^{-1/2}du=\sqrt{u}+C=\sqrt{x^2-2x}+C$$

Answer 3

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 $c\in\mathbb{R}$.

Answer 4

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

Answer 5

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)