Question

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

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. $\frac{4}{3}$

Answer 1

The correct option is B $\frac{2}{3}$

$I=\underset{0}{\overset{\infty }{\int }}\frac{1}{(x+\sqrt{1+x^2}{)}^2}dx$

Let $x=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta$

$I=\underset{0}{\overset{\pi /2}{\int }}\frac{{\sec }^2\theta }{(\tan \theta +\sec \theta {)}^2}d\theta$
$=\underset{0}{\overset{\pi /2}{\int }}{\sec }^2\theta (\sec \theta -\tan \theta {)}^{2}d\theta$

Let $\sec \theta -\tan \theta =t$
$\Rightarrow \sec \theta (\tan \theta -\sec \theta )d\theta =dt$

$I=\underset{1}{\overset{0}{\int }}\frac{1}{2}(t+\frac{1}{t})t(-dt)$
$=\underset{0}{\overset{1}{\int }}\frac{1}{2}(t^2+1)dt$
$=\frac{1}{2}{[\frac{t^3}{3}+t]}_0^1$
$=\frac{2}{3}$


ALTERNATE SOLUTION:
$\frac{1}{x+\sqrt{1+x^2}}=\sqrt{1+x^2}-x$

Let $\sqrt{1+x^2}-x=t$
$\Rightarrow \sqrt{1+x^2}=x+t$
Squaring both sides
$1+x^2=x^2+2xt+t^2\Rightarrow x=\frac{1}{2t}-\frac{t}{2}=\frac{1-t^2}{2t}dx=(\frac{-1}{2t^2}-\frac{1}{2})dt=-\left(\frac{1+t^2}{2t^2}\right)dt$
When $x=0\Rightarrow t=1x=\infty \Rightarrow t=0$

$\therefore I=1-t^2\underset{1}{\overset{0}{\int }}\frac{-\left(\frac{1+t^2}{2t^2}\right)}{(\frac{1-t^2}{2t}+\frac{1-t^2}{2t}+t)}dt$
$I=1-t^2\underset{0}{\overset{1}{\int }}\frac{\left(\frac{1+t^2}{2t^2}\right)}{\frac{1}{t^2}}dt$
$I=\frac{1}{2}\underset{0}{\overset{1}{\int }}(1+t^2)dt$
$=\frac{1}{2}{[t+\frac{t^3}{3}]}_0^1$
$=\frac{2}{3}$