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

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

• $$\dfrac{d(x^2-2x)}{dx}=?$$ – lab bhattacharjee Dec 31 '15 at 12:41
• hint $x^2-2x=(x-1)^2-1$ could give you an idea. – Claude Leibovici Dec 31 '15 at 12:41
• Over-eager answers should have waited for Matt to respond to either of these fine hints. – GEdgar Dec 31 '15 at 13:24

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

• There appears to be a typo at your last line after the first equality sign. – Nigel Overmars Dec 31 '15 at 12:57

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

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

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

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 $c\in\mathbb{R}$.

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

• I see absolutely no reason for downvoting this answer. – egreg Dec 31 '15 at 13:45
• See @GEdgar's comment. Although I am not the downvoter, but there's your reason. – Ron Gordon Dec 31 '15 at 15:33

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)

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,

# $$I=\frac{1}{2} \frac{t^\frac{1}{2}}{\frac{1}{2}} + C$$

now substitute the value of t,

# $$I=\sqrt(x^2-2x) + C-------(2)$$

equation (2) is the required integration of equation (1)