Question

If ${\int }_{log2}^a\frac{e^x}{\sqrt{e^x-1}}dx=2$, then $a=$ ________.

• A. $ln5$
• B. $2\log 2$
• C. $\log 2$
• D. $0$

Answer 1

The correct option is A $ln5$

Let $t=\sqrt{e^x-1}$

$⟹dt=\frac{e^x}{2\sqrt{e^x-1}}dx$

$⟹\frac{e^x}{\sqrt{e^x-1}}dx=2dt$

${\int }_{ln2}^a\frac{e^x}{\sqrt{e^x-1}}dx$

$={\int }_1^{\sqrt{e^a-1}}2dt$

$=2(\sqrt{e^a-1}-1)$

$⟹2(\sqrt{e^a-1}-1)=2$

$⟹\sqrt{e^a-1}=2$

$⟹e^a=5$

$⟹a=ln\left(5\right)$

Hence, the answer is option (A).