Question

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

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{8}$

Answer 1

The correct option is D $\frac{\pi }{8}$

Let $I={\int }_0^{\infty }\frac{x{\tan }^{-1}x}{(1+x^2)}dx$

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

$\therefore I={\int }_0^{\pi /2}\frac{\theta \tan \theta }{{\sec }^2\theta }d\theta ={\int }_0^{\pi /2}\theta \sin \theta \cos \theta d\theta$

$=\frac{1}{2}{\int }_0^{\pi /2}\theta \sin 2\theta d\theta$

$=\frac{1}{2}{[-\theta \frac{\cos 2\theta }{2}+\int \frac{\cos 2\theta }{2}d\theta ]}_{{}^{\pi /2}}^0=\frac{\pi }{8}$