Question

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

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

Answer 1

The correct option is C $\log 4$

Let $I={\int }_{\log 2}^x\frac{1}{\sqrt{e^x-1}}dx$
Put $e^x-1=t^2$ $\Rightarrow e^{x}dx=2tdt\Rightarrow (t^2+1)dx=2tdt\Rightarrow dx=\frac{2t}{1+t^2}dt$

When $x=\log 2,t=\pm 1$

$\therefore I={\int }_1^{\sqrt{e^x-1}}\frac{2t}{t(1+t^2)}dt$

$=2{\int }_1^{\sqrt{e^x-1}}\frac{1}{1+t^2}dt$

$=2|{\tan }^{-1}t{|}_1^{\sqrt{e^x-1}}$

$=2[{\tan }^{-1}\sqrt{e^x-1}-{\tan }^{-1}1]$

$=2{\tan }^{-1}\sqrt{e^x-1}-\frac{\pi }{2}$
But $I={\int }_{\log 2}^x\frac{1}{\sqrt{e^x-1}}dx=\frac{\pi }{6}$

$\Rightarrow 2{\tan }^{-1}\sqrt{e^x-1}-\frac{\pi }{2}=\frac{\pi }{6}$
$\Rightarrow 2{\tan }^{-1}\sqrt{e^x-1}=\frac{2\pi }{3}$
$\Rightarrow \sqrt{e^x-1}=\tan \left(\frac{\pi }{3}\right)=\sqrt{3}$
$\Rightarrow e^x-1=3$
$\Rightarrow e^x=4$
$\therefore x=\log 4$