Question

If $\underset{\log 2}{\overset{x}{\int }}\frac{1}{\sqrt{e^x-1}}dx=\frac{\pi }{6}$, then the value of $x$ is

• A. log 2
• B. log 3
• C. log 4
• D. None of these

Answer 1

Answer: (C)

Let $l=\underset{\log 2}{\overset{x}{\int }}\frac{1}{\sqrt{e^x-1}}dx$
Put $e^x-1=t^2$
$\Rightarrow e^{x}dx=2tdt$
$\therefore l=2\underset{1}{\overset{\sqrt{e^x-1}}{\int }}\frac{1}{1+t^2}dt$
$=2\left[{\tan }^{-1}\sqrt{e^x-1}-{\tan }^{-1}1\right]$
$=2{\tan }^{-1}\sqrt{e^x-1}-\frac{\pi }{4}$
But $\underset{{\log }_0^2}{\overset{x}{\int }}\frac{1}{\sqrt{e^x-1}}dx=\frac{\pi }{6}$
$=2{\tan }^{-1}\sqrt{e^x-1}$
$=\frac{2\pi }{3}$
$\Rightarrow \sqrt{e^x-1}=\tan \frac{\pi }{3}$
$=\sqrt{3}$
$=e^x-1$
$=3$
$\Rightarrow e^x=4$
$\therefore x=\log 4$