Question

If $ \int\limits^{x}_{\log \, 2} \frac{du}{(e^u - 1)^{1/2}} = \frac{\pi}{6},$ then $e^x$ is equal to :

• A. 1
• B. 2
• C. 4
• D. -1

Answer 1

Answer: (C)

Let $\int\limits_{\log 2}^{x} \frac{d u}{\left(e^{u}-1\right)^{1 / 2}}$
or $I=\int\limits_{\log 2}^{x} \frac{e^{u}}{e^{u}\left(e^{u}-1\right)^{1 / 2}} d u$
Let $e^{u}-1=t^{2} \Rightarrow e^{u} d u=2 t d t$
$=\int\limits_{1}^{\sqrt{e^{x}}-1} \frac{2 t}{\left(t^{2}+1\right) t} d t=2 \int\limits_{1}^{\sqrt{e^{x}}-1} \frac{d t}{\left(1+t^{2}\right)}$
$\left.=\left[\tan ^{-1} t\right]_{1}^{\sqrt{e^{x}-1}}=2 \tan ^{-1} \sqrt{e^{x}-1}-\tan ^{-1} 1\right]$
$\Rightarrow 2\left[\tan ^{-1} \sqrt{e^{x}-1}-\frac{\pi}{4}\right]=\frac{\pi}{6}$ (given)
$\Rightarrow \tan ^{-1} \sqrt{e^{x}-1}=\frac{\pi}{12}+\frac{\pi}{4}=\frac{\pi}{3}$
$\Rightarrow \sqrt{e^{x}-1}=\tan \left(\frac{\pi}{3}\right)$
$\sqrt{e^{x}-1}=\sqrt{3}$
$\Rightarrow e^{x}=3+1=4$