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Integral Question - $\displaystyle\int\frac{1}{\sqrt{x^2-x}}\,\mathrm dx$. $$\int\frac{1}{\sqrt{x(x-1)}}\,\mathrm dx =\int \left(\frac{A}{\sqrt x} + \frac{B}{\sqrt{x-1}}\right)\,\mathrm dx$$
This is the right way to solve it?

Thanks!

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  • $\begingroup$ There are no constants $A,B$ such that $$\frac{1}{\sqrt{x(x-1)}}=\frac{A}{\sqrt x} + \frac{B}{\sqrt{x-1}}$$ $\endgroup$ – Américo Tavares May 1 '13 at 10:04
  • $\begingroup$ @Ofir : No. $\sqrt{CD} \neq \sqrt{C} + \sqrt{D}$. Complete the square inside the square root. $\endgroup$ – Stefan Smith May 1 '13 at 10:57
  • $\begingroup$ Yes, I understand that after lab told me, thanks $\endgroup$ – Ofir Attia May 1 '13 at 10:59
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The Partial Fraction Decomposition is for rational fraction only.

$$\int\frac{dx}{\sqrt{x^2-x}}=\int\frac{2dx}{\sqrt{4x^2-4x}}=\int\frac{2dx}{\sqrt{(2x-1)^2-1^2}}$$

Now, put $2x-1=\sec\theta$

EDIT: completing as requested

So,$2dx=\sec\theta\tan\theta d\theta$

$$\text{So,}\int\frac{2dx}{\sqrt{(2x-1)^2-1^2}}=\int \frac{\sec\theta\tan\theta d\theta}{\tan\theta}=\int \sec\theta d\theta =\ln|\sec\theta+\tan\theta|+C $$ (where $C$ is an arbitrary constant of indefinite integral )

$$=\ln\left|2x-1+\sqrt{(2x-1)^2-1}\right|+C=\ln\left|2x-1+2\sqrt{x^2-x}\right|+C$$


Alternatively,using $$\frac{dy}{\sqrt{y^2-a^2}}=\ln\left|y+\sqrt{y^2-a^2}\right|+C$$

$$\int\frac{dx}{\sqrt{x^2-x}}=\int\frac{dx}{\sqrt{\left(x-\frac12\right)^2-\left(\frac12\right)^2}}$$ $$=\ln\left|x-\frac12+\sqrt{x^2-x}\right|+C=\ln\left|2x-1+2\sqrt{x^2-x}\right|+C-\ln2=\ln\left|2x-1+2\sqrt{x^2-x}\right|+C'$$ where $C'=C-\ln2$ another arbitrary constant

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  • $\begingroup$ ok thanks, got it! $\endgroup$ – Ofir Attia May 1 '13 at 9:53
  • $\begingroup$ @OfirAttia, can $1$ be equal to $A\sqrt{x-1}+B\sqrt x?$ $\endgroup$ – lab bhattacharjee May 1 '13 at 9:54
  • $\begingroup$ no, I see it right now. thanks $\endgroup$ – Ofir Attia May 1 '13 at 9:55
  • $\begingroup$ I can put outside the intergral 1/2? because the 2dx? $\endgroup$ – Ofir Attia May 1 '13 at 9:57
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    $\begingroup$ @OfirAttia,$\int\frac{dx}{\sqrt{x-x^2}}=\int\frac{dx}{\sqrt{\left(\frac12\right)^2-\left(x-\frac12\right)^2}}$ Use $$\int \frac{dy}{\sqrt{a^2-y^2}}=\arcsin \frac ya$$ $\endgroup$ – lab bhattacharjee May 2 '13 at 5:15

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