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

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

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    $\begingroup$ $$\dfrac{d(x^2-2x)}{dx}=?$$ $\endgroup$ – lab bhattacharjee Dec 31 '15 at 12:41
  • $\begingroup$ hint $x^2-2x=(x-1)^2-1$ could give you an idea. $\endgroup$ – Claude Leibovici Dec 31 '15 at 12:41
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    $\begingroup$ Over-eager answers should have waited for Matt to respond to either of these fine hints. $\endgroup$ – GEdgar Dec 31 '15 at 13:24
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$$t=x^2-2x$$ $$dt=2x-2dx$$ $$dx=\frac{dt}{2(x-1)}$$ $$\int\frac{x-1}{\sqrt{x^2-2x}}dx=\int\frac{1}{2}\frac{1}{t}dt=\frac{1}{2}\int t^{\frac{-1}{2}}=\frac{1}{2}2t^{\frac{1}{2}} +C = \sqrt{x^2-2x}+C$$

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    $\begingroup$ There appears to be a typo at your last line after the first equality sign. $\endgroup$ – Nigel Overmars Dec 31 '15 at 12:57
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$$\int\frac{x-1}{\sqrt{ x^2-2x}}dx=\frac12\int\frac{2x-2}{\sqrt{ x^2-2x}}dx$$

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

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

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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 $c\in\mathbb{R}$.

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

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

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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)

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