# Integral Question - $\int\frac{1}{\sqrt{x^2-x}}\,\mathrm dx$

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!

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

## 1 Answer

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

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