Question

${\int }_0^{\infty }\frac{dx}{\left(x+{\left(\sqrt{x^2+1}\right)}^3\right)}=$

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

Answer 1

The correct option is A

$\frac{3}{8}$

Explanation for Correct answer:

$\text{I}={\int }_0^{\infty }\frac{dx}{{\left(x+\left(\sqrt{x^2+1}\right)\right)}^3}$

Using the substitution method to solve, we get

$$\begin{gathered} x=\tan \theta \\ \Rightarrow dx=se{c}^2\theta .d\theta \end{gathered}$$

limits

$x$	$0$	$\infty$
$\theta$	$$\begin{gathered} \theta ={\tan }^{-1}\left(0\right) \\ =0 \end{gathered}$$	$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{1}{0}\right) \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$

${\int }_0^{\infty }\frac{dx}{{\left(x+\left(\sqrt{x^2+1}\right)\right)}^3}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{se{c}^2\theta .d\theta }{{\left(\tan \theta +\left(\sqrt{{\tan }^2\theta +1}\right)\right)}^3}$

$\begin{array}{rcl}& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{se{c}^2\theta .d\theta }{{\left(\tan \theta +\sqrt{se{c}^2\theta }\right)}^3}\left[\because se{c}^2\theta ={\tan }^2\theta +1\right]\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{se{c}^2\theta .d\theta }{{\left(\tan \theta +sec\theta \right)}^3}\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{se{c}^2\theta .d\theta }{{\left({\displaystyle \frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }}\right)}^3}\left[\because \tan \theta =\frac{\sin \theta }{\cos \theta };sec\theta =\frac{1}{\cos \theta }\right]\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{se{c}^2\theta .d\theta }{\frac{1}{{\cos }^3\theta }{\left({\displaystyle 1+\sin \theta }\right)}^3}\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{se{c}^2\theta .d\theta }{se{c}^3\theta {\left({\displaystyle 1+\sin \theta }\right)}^3}\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{d\theta }{sec\theta {\left({\displaystyle 1+\sin \theta }\right)}^3}\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{d\theta }{sec\theta {\left({\displaystyle 1+\sin \theta }\right)}^3}\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos \theta d\theta }{{\left({\displaystyle 1+\sin \theta }\right)}^3}\left[\because sec\theta =\frac{1}{\cos \theta }\right]\end{array}$

Using the substitution method to solve, we get

$$\begin{gathered} t=1+\sin \theta \\ \Rightarrow dt=\cos \theta .d\theta \end{gathered}$$

limits

$\theta$	$0$	$\frac{\mathrm{\pi }}{2}$
$t$	$$\begin{gathered} t=1+\sin \left(0\right) \\ =1+0 \\ =1 \end{gathered}$$	$$\begin{gathered} t=1+\sin \left(\frac{\mathrm{\pi }}{2}\right) \\ =1+1 \\ =2 \end{gathered}$$

$$\begin{gathered} ={\int }_1^2\frac{dt}{{{\displaystyle t}}^3} \\ =\frac{-1}{2}{\left[t^{-2}\right]}_1^2 \\ =\frac{-1}{2}\left[2^{-2}-1^{-2}\right] \\ =\frac{-1}{2}\left[\frac{1}{4}-1\right] \\ =\frac{-1}{2}\left[\frac{-3}{4}\right] \\ =\frac{3}{8} \end{gathered}$$

Hence, option (A) is the correct answer.