Question

${\int }_0^{\infty }\frac{dx}{(x+\sqrt{x^2+1}{)}^5}=$

• A. $\frac{1}{5}$
• B. $\frac{4}{25}$
• C. $\frac{5}{24}$
• D. $\frac{1}{6}$

Answer 1

The correct option is C $\frac{5}{24}$

Let
$x=\tan \theta \Rightarrow I={\int }_0^{\pi /2}\frac{{\sec }^2\theta d\theta }{(\tan \theta +\sec \theta {)}^5}\sec \theta +\tan \theta =t\Rightarrow \sec \theta -\tan \theta =\frac{1}{t}\Rightarrow \sec \theta =\frac{1}{2}(t+\frac{1}{t})$
Now $dt=\sec \theta (\sec \theta +\tan \theta )d\theta$
$\frac{dt}{t}=\sec \theta d\theta$

$\therefore I=\frac{1}{2}{\int }_1^{\infty }\frac{(t+1/t)}{t^6}dt=\frac{1}{2}{\int }_1^{\infty }(\frac{1}{t^5}+\frac{1}{t^7})dt=\frac{5}{24}$