Question

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

• A. 1
• B. 2
• C. 4
• D. -1

Answer 1

Answer: (C)

Let $\underset{\log 2}{\overset{x}{\int }}\frac{du}{{\left(e^u-1\right)}^{1/2}}$
or $I=\underset{\log 2}{\overset{x}{\int }}\frac{e^u}{e^u{\left(e^u-1\right)}^{1/2}}du$
Let $e^u-1=t^2\Rightarrow e^{u}du=2tdt$
$=\underset{1}{\overset{\sqrt{e^x}-1}{\int }}\frac{2t}{\left(t^2+1\right)t}dt=2\underset{1}{\overset{\sqrt{e^x}-1}{\int }}\frac{dt}{\left(1+t^2\right)}$
$={\left[{\tan }^{-1}t\right]}_1^{\sqrt{e^x-1}}=2{\tan }^{-1}\sqrt{e^x-1}-{\tan }^{-1}1]$
$\Rightarrow 2\left[{\tan }^{-1}\sqrt{e^x-1}-\frac{\pi }{4}\right]=\frac{\pi }{6}$ (given)
$\Rightarrow {\tan }^{-1}\sqrt{e^x-1}=\frac{\pi }{12}+\frac{\pi }{4}=\frac{\pi }{3}$
$\Rightarrow \sqrt{e^x-1}=\tan \left(\frac{\pi }{3}\right)$
$\sqrt{e^x-1}=\sqrt{3}$
$\Rightarrow e^x=3+1=4$