Question

Integrate:

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

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

Answer 1

The correct option is D $\pi /8$

We have, ${\int }_0^{\infty }\frac{x{\tan }^{-1}x}{(1+x^2{)}^2}dx...\left(1\right)$

Let $ta{n}^{-1}x=z\Rightarrow x=tanz$

$\Rightarrow \frac{dx}{1+x^2}=dz$

When $x=0$ then $z=0$

and when $x=\infty$ then $z=\frac{\pi }{2}$

Now, $\left(1\right)$ becomes

${\int }_0^{\frac{\pi }{2}}\frac{ztanz}{(1+ta{n}^{2}z)}dz$

$={\int }_0^{\frac{\pi }{2}}zsinzcoszdz$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}z\left(2sinzcosz\right)dz$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}zsin\left(2z\right)dz[\because 2sinAcosA=sin2A]...\left(2\right)$

Let $2z=t$

$\Rightarrow 2dz=dt$

$\Rightarrow dz=\frac{1}{2}dt$

When $z=0$ then $t=0$

and when $z=\frac{\pi }{2}$ then $t=\pi$

Now, $\left(2\right)$ becomes,

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

$=\frac{1}{8}{\int }_0^{\pi }tsin\left(t\right)dt$

$=\frac{1}{8}[t\int sin(t)dt{]}_0^{\pi }-\frac{1}{8}{\int }_0^{\pi }[\frac{d}{dt}.t\int sin\left(t\right)]dt$

$=\frac{1}{8}[-tcos(t){]}_0^{\pi }+\frac{1}{8}{\int }_0^{\pi }cos(t)dt$

$=\frac{1}{8}[-\pi cos(\pi )+0]+\frac{1}{8}\left[sin\right(t){]}_0^{\pi }$

$=\frac{1}{8}\pi +\frac{1}{8}(sin\pi -sin0)$

$=\frac{\pi }{8}+0$

$=\frac{\pi }{8}$

So, $\text{D}$ is the correct option.