Question

$\underset{\log 2}{\overset{a}{\int }}\frac{e^x}{\sqrt{e^x-1}}dx=2,$ then $a=$

• A. $0$
• B. $2\log 2$
• C. $\log 5$
• D. $\log 2$

Answer 1

The correct option is C $\log 5$

$I=\underset{\log 2}{\overset{a}{\int }}\frac{e^x}{\sqrt{e^x-1}}dx$
Let $e^x-1=t\Rightarrow e^{x}dx=dt$
$I=\underset{1}{\overset{e^a-1}{\int }}\frac{1}{\sqrt{t}}dt$
$\therefore {\left[2\sqrt{t}\right]}_1^{e^a-1}=2$

$\Rightarrow 2\sqrt{e^a-1}-2=2\Rightarrow e^a=5\Rightarrow a={\log }_{e}5$