Question

If $\underset{\ln 2}{\overset{k}{\int }}\frac{1}{\sqrt{e^x-1}}dx=\frac{\pi }{6},$ then $k=\ln p.$ The value of $p+1$ is

Answer 1

$\underset{\ln 2}{\overset{k}{\int }}\frac{1}{\sqrt{e^x-1}}dx$
Let $e^x-1=t^2$
$\Rightarrow e^{x}dx=2tdt$
$\Rightarrow dx=\frac{2t}{t^2+1}dt$

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

Now, $\underset{\ln 2}{\overset{k}{\int }}\frac{1}{\sqrt{e^x-1}}dx=\frac{\pi }{6}$
$\Rightarrow 2{\tan }^{-1}\sqrt{e^k-1}-2{\tan }^{-1}\sqrt{e^{\ln 2}-1}=\frac{\pi }{6}$
$\Rightarrow 2{\tan }^{-1}\sqrt{e^k-1}-2{\tan }^{-1}1=\frac{\pi }{6}$
$\Rightarrow 2{\tan }^{-1}\sqrt{e^k-1}-\frac{\pi }{2}=\frac{\pi }{6}$
$\Rightarrow {\tan }^{-1}\sqrt{e^k-1}=\frac{\pi }{3}$
$\Rightarrow \sqrt{e^k-1}=\sqrt{3}$
$\Rightarrow e^k=4$
$\Rightarrow k=\ln 4$
Thus $p=4$
$\Rightarrow p+1=5$