Question

Consider $I_n=\underset{0}{\overset{\infty }{\int }}\frac{dx}{{(x+\sqrt{1+x^2})}^n},$ where $n>1$, then which of the following statement(s) is/are true ?

• A. $I_2=\frac{2}{3}$
• B. $I_2=\frac{2}{5}$
• C. $I_{10}<I_5$
• D. $I_{10}>I_5$

Answer 1

Answer: (A), (C)

The correct options are
A $I_2=\frac{2}{3}$
C $I_{10}<I_5$

Let $x+\sqrt{1+x^2}=t$
$\Rightarrow 1+x^2=t^2+x^2-2tx$
$x=\frac{t^2-1}{2t}$
$I_n=\underset{1}{\overset{\infty }{\int }}\frac{1}{t^n}\left(\frac{t+\frac{1}{t}}{2t}\right)dt$
$=\frac{1}{2}\underset{1}{\overset{\infty }{\int }}\frac{t^2+1}{t^{n+2}}dt$
$=\frac{1}{2}{[\frac{t^{1-n}}{1-n}+\frac{t^{-n-1}}{-n-1}]}_1^{\infty }=\frac{n}{n^2-1}$
$I_2=\frac{2}{3}$ $I_{10}=\frac{10}{99},I_5=\frac{5}{24}$
$I_{10}<I_5$