Question

${\int }_0^{\infty }\frac{dx}{{[x+\sqrt{x^2+1}]}^3}$ is equal to

• A. $\frac{3}{8}$
• B. $\frac{1}{8}$
• C. $-\frac{3}{8}$
• D. none of these

Answer 1

The correct option is A $\frac{3}{8}$

Putting $x=\tan \theta$,we get
${\int }_0^{\infty }\frac{dx}{{[x+\sqrt{x^2+1}]}^3}={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^2\theta d\theta }{(\tan \theta +\sec \theta {)}^3}$
$={\int }_0^{\frac{\pi }{2}}\frac{\cos \theta }{(1+\sin \theta {)}^3}d\theta$
$={\left[\frac{1}{2(1+\sin \theta {)}^2}\right]}_0^{\frac{\pi }{2}}=-\frac{1}{8}+\frac{1}{2}=\frac{3}{8}$