Question

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

• A. $\frac{\pi }{2}-\log 2$
• B. $\frac{\pi }{2}+\log 2$
• C. $\frac{\pi }{3}-\log 3$
• D. $\frac{\pi }{2}-2\log 2$

Answer 1

Answer: (A)

$\frac{\pi }{2}-\log 2$

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

Let $x=\tan \theta$ $⟹dx={\sec }^2\theta d\theta$

Now, when $x=0,\theta =0$ and when $x=1,\theta =\frac{\pi }{4}$

Again, $\frac{2x}{1-x^2}=\frac{2\tan \theta }{1-{\tan }^2\theta }=\tan 2\theta$

And ${\tan }^{-1}\frac{2x}{1-x^2}={\tan }^{-1}\left(\tan 2\theta \right)=2\theta$

Hence, integration becomes

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

$=2{[\theta \tan \theta -\log |\sec \theta |]}_0^{\pi /4}$

$=2[\frac{\pi }{4}-\log \sqrt{2}]$

$=2[\frac{\pi }{4}-\frac{1}{2}\log 2]$

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