Question

Value of ${\int }_{-1}^1\frac{x}{\sqrt{1-x^2}}\log \frac{1+x}{1-x}dx$ is

• A. $-\pi$
• B. $2\pi$
• C. $2$
• D. $-2$

Answer 1

The correct option is B $2\pi$

${\int }_{-1}^1\frac{x}{\sqrt{1-x^2}}\log \frac{1+x}{1-x}dx$
$={\left[\log \left(\frac{1+x}{1-x}\right)(-\sqrt{1-x^2})\right]}_{-1}^1-{\int }_{-1}^1(-\sqrt{1-x^2})(-\frac{1}{x^2-1})dx$
$={\int }_{-1}^1\frac{2}{\sqrt{1-x^2}}dx={\left[2{\sin }^{-1}x\right]}_{-1}^1=\left[2(\frac{\pi }{2}+\frac{\pi }{2})\right]=2\pi$