Question

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

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

Answer 1

Answer: (D)

Put $x = tan\, z$, $dx = sec^{2} z \, dz$
when $x = 0$ ; when $x=\infty, z=\pi /2$
$\therefore $ given integral
$=\int\limits_{0}^{\pi /2} \frac{\left(tan\, z\right)}{\left(1+tan^{2}\,z\right)^{2}} sec^{2} \, z \, dz$
$=\int\limits_{0}^{\pi /2} \frac{z\, tan \,z}{sec^{2}\,z} dz$
$=\int\limits^{\pi /2}_{0} z \frac{sin\,z}{cos \, z} cos^{2} z\, dz$
$=\int\limits_{0}^{\pi /2} z\, sin \, z\, cos\, z\, dz$
$=\frac{1}{2} \int\limits_{0}^{\pi /2}z\left(sin\,2z\right)dz$
$=\frac{1}{2}\left[\left|z-\left(-\frac{cos\,2z}{2}\right)\right|_{0}^{\pi /2}-\int_{0}^{\pi /2}\left(1\right)\left(-\frac{cos\,2z}{2}\right)dz\right]$
$=\frac{1}{2}\left[-\frac{\pi}{4}cos\,\pi+\frac{1}{2} \left|\frac{sin\, 2z}{2}\right|_{0}^{\pi /2}\right]$
$=\frac{1}{2} \left[\frac{\pi}{4}+0\right]$
$=\frac{\pi}{8}$