Question

The value of integral $\underset{-1/\sqrt{3}}{\overset{1/\sqrt{3}}{\int }}\frac{x^4}{1-x^4}\cdot {\cos }^{-1}\left(\frac{2x}{1-x^2}\right)dx$ is

• A. $\frac{\pi }{12}[\pi +3\ln (2+\sqrt{3})-4\sqrt{3}]$
• B. $\frac{\pi }{6}[\pi +\ln (2+\sqrt{3})-2\sqrt{3}]$
• C. $\frac{\pi }{4}[\pi +\ln (2+\sqrt{3})+\sqrt{3}]$
• D. $\frac{\pi }{3}[\pi +2\ln (2+\sqrt{3})+3\sqrt{3}]$

Answer 1

The correct option is A $\frac{\pi }{12}[\pi +3\ln (2+\sqrt{3})-4\sqrt{3}]$

Let
$I=\underset{-1/\sqrt{3}}{\overset{1/\sqrt{3}}{\int }}\frac{x^4}{1-x^4}\cdot {\cos }^{-1}\left(\frac{2x}{1-x^2}\right)dx=\underset{-1/\sqrt{3}}{\overset{1/\sqrt{3}}{\int }}\frac{x^4}{1-x^4}\cdot [\frac{\pi }{2}-{\sin }^{-1}\left(\frac{2x}{1-x^2}\right)]dx=\frac{\pi }{2}\underset{-1/\sqrt{3}}{\overset{1/\sqrt{3}}{\int }}\frac{x^4}{1-x^4}dx-\underset{-1/\sqrt{3}}{\overset{1/\sqrt{3}}{\int }}\frac{x^4}{1-x^4}\cdot {\sin }^{-1}\left(\frac{2x}{1-x^2}\right)dx$
We know that,
$\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\{\begin{array}{c}2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\text{if}f\left(x\right)\text{is even}\\ 0,\text{if}f\left(x\right)\text{is odd}\end{array}$
$\Rightarrow I=\pi \underset{0}{\overset{1/\sqrt{3}}{\int }}\frac{x^4}{1-x^4}dx-0$
$\Rightarrow I=\frac{\pi }{2}\underset{0}{\overset{1/\sqrt{3}}{\int }}(-2+\frac{1}{1-x^2}+\frac{1}{1+x^2})dx=\frac{\pi }{2}{[-2x+\frac{1}{2}\ln \left|\frac{1+x}{1-x}\right|+{\tan }^{-1}x]}_0^{1/\sqrt{3}}=\frac{\pi }{12}[\pi +3\ln (2+\sqrt{3})-4\sqrt{3}]$