Question

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

• A. $\pi /2$
• B. $\pi /6$
• C. $\pi /4$
• D. $\pi /8$

Answer 1

The correct option is D $\pi /8$

Let, ${\tan }^{-1}x=t⟹\frac{1}{1+x^2}dx=dt$

As $x\to 0⟹t\to 0$

As $x\to \infty ⟹t\to \pi /2$

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

$={\int }_0^{\pi /2}\frac{\tan t.\left(t\right)}{1+{\tan }^{2}t}dt$

$={\int }_0^{\pi /2}\frac{\tan t.\left(t\right)}{{\sec }^{2}t}dt$

$={\int }_0^{\pi /2}\sin t\cos t.\left(t\right)dt$

$=\frac{1}{2}{\int }_0^{\pi /2}2\sin t\cos t.\left(t\right)dt$

$=\frac{1}{2}{\int }_0^{\pi /2}\sin 2t.\left(t\right)dt$

$=\frac{1}{2}{[-t.\frac{\cos 2t}{2}+\frac{\sin 2t}{4}]}_0^{\pi /2}$

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

$=\frac{\pi }{8}$ (Ans)