Question

If ${\int }_{ln2}^x\frac{du}{(e^u-1{)}^{\frac{1}{2}}}=\frac{\pi }{6}$, then $e^x$ equals

• A. 1
• B. 2
• C. 4
• D. 3

Answer 1

The correct option is C

4

${\int }_{ln2}^x\frac{du}{(e^u-1{)}^{\frac{1}{2}}}=\frac{\pi }{6}$

Put $e^u-1=t^2$

$\Rightarrow {\int }_1^{\sqrt{e^x-1}}\frac{2t}{t(1+t^2)}dt=\frac{\pi }{6}$

$\Rightarrow 2(ta{n}^{-1}t{)}_1^{\sqrt{e^x-1}}=\frac{\pi }{6}$

$\Rightarrow ta{n}^{-1}\sqrt{e^x-1}-\frac{\pi }{4}=\frac{\pi }{12}$

$\Rightarrow \sqrt{e^x-1}=tan\frac{\pi }{3}\Rightarrow \sqrt{e^x-1}=\sqrt{3}\Rightarrow e^x=4$