Question

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

then $x$ is equal to?

• A. $e^2$
• B. $1/e$
• C. $\log 4$
• D. $\log 2$

Answer 1

The correct option is C $\log 4$

Let $I=\int \frac{1}{\sqrt{e^x-1}}dx$
Substituting $t=e^x\Rightarrow dt=e^{x}dx$, we get
$I=\int \frac{1}{t\sqrt{t-1}}dt$
Again substituting $u=t-1\Rightarrow du=dt$, we get
$I=\int \frac{1}{\sqrt{u}(u+1)}du$
now substituting $s=\sqrt{u}\Rightarrow ds=\frac{1}{2\sqrt{u}}du$, we get
$I=2\int \frac{1}{s^2+1}ds=2ta{n}^{-1}s=2ta{n}^{-1}\sqrt{u}=2ta{n}^{-1}\sqrt{t-1}=2ta{n}^{-1}\sqrt{e^x-1}$
Therefore,
${\int }_{\log 2}^{x}2ta{n}^{-1}\left(\sqrt{e^x-1}\right)=\frac{\pi }{6}\Rightarrow {\left[2ta{n}^{-1}\left(\sqrt{e^x-1}\right)\right]}_{\log 2}^x=\frac{\pi }{6}$
$\therefore x=\log 4$