Question

The value of $\underset{0}{\overset{\infty }{\int }}\frac{x{\tan }^{-1}x}{(1+x^2{)}^2}$ dx is

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

Answer 1

Answer: (D)

Put $x=tanz$, $dx=se{c}^{2}zdz$
when $x=0$ ; when $x=\infty ,z=\pi /2$
$\therefore$ given integral
$=\underset{0}{\overset{\pi /2}{\int }}\frac{\left(tanz\right)}{{\left(1+ta{n}^{2}z\right)}^2}se{c}^{2}zdz$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{ztanz}{se{c}^{2}z}dz$
$=\underset{0}{\overset{\pi /2}{\int }}z\frac{sinz}{cosz}co{s}^{2}zdz$
$=\underset{0}{\overset{\pi /2}{\int }}zsinzcoszdz$
$=\frac{1}{2}\underset{0}{\overset{\pi /2}{\int }}z\left(sin2z\right)dz$
$=\frac{1}{2}\left[{\left|z-\left(-\frac{cos2z}{2}\right)\right|}_0^{\pi /2}-{\int }_0^{\pi /2}\left(1\right)\left(-\frac{cos2z}{2}\right)dz\right]$
$=\frac{1}{2}\left[-\frac{\pi }{4}cos\pi +\frac{1}{2}{\left|\frac{sin2z}{2}\right|}_0^{\pi /2}\right]$
$=\frac{1}{2}\left[\frac{\pi }{4}+0\right]$
$=\frac{\pi }{8}$