Question

${\int }_0^1{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)dx$

• A. $\frac{\pi }{2}-\ln 2$
• B. $\frac{\pi }{2}-\ln 4$
  
• C. $\frac{\pi }{4}-\ln 2$
  
• D. $\frac{\pi }{4}-\ln 4$
  

Answer 1

The correct option is A $\frac{\pi }{2}-\ln 2$

${\int }_0^1{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)dx$

Putting $x=\tan \theta$ $\Rightarrow dx={\sec }^2\theta d\theta$

$={\int }_0^{\frac{\pi }{4}}{\tan }^{-1}\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right){\sec }^2\theta d\theta$

$={\int }_0^{\frac{\pi }{4}}{\tan }^{-1}\left(\tan 2\theta \right){\sec }^2\theta d\theta$

$={\int }_0^{\frac{\pi }{4}}2\theta {\sec }^2\theta d\theta$

$=2{\int }_0^{\frac{\pi }{4}}\theta {\sec }^2\theta d\theta$

$=2\left[{\left(\theta \tan \theta \right)}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}1.\tan \theta d\theta \right]$

$=2{\left[\theta \tan \theta -\ln \left|\sec \theta \right|\right]}_0^{\frac{\pi }{4}}$

$=2{\left[\theta \tan \theta +\ln \left|\cos \theta \right|\right]}_0^{\frac{\pi }{4}}$

$=2\left[\left(\frac{\pi }{4}+\ln \frac{1}{\sqrt{2}}\right)-0\right]$

$=2\left[\frac{\pi }{4}+\ln \frac{1}{\sqrt{2}}\right]$

$=\frac{2\pi }{4}+2\ln {\left(2\right)}^{\frac{-1}{2}}$

$=\frac{\pi }{2}+2\times \frac{-1}{2}ln2$

$=\frac{\pi }{2}-\ln 2$