Question

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

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

Answer 1

The correct option is C $\pi /4$

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

$subtituting,{tan}^{-1}x=t$

$\Rightarrow x=tant$

$\Rightarrow dx={sec}^{2}t$

now I becomes,

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{t\times tant\times {sec}^{2}t}{{sec}^{4}t}dt$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{tsint}{cost}{cos}^{2}tdt$

$\Rightarrow I=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}tsin2tdt$

$usingbyparts,$

$\Rightarrow I=\frac{1}{2}{\left[t\{-\frac{cos2t}{2}\}\right]}_0^{\frac{\pi }{2}}-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\left[1\{-\frac{cos2t}{2}\}\right]dt$

$\therefore I=\frac{\pi }{4}$