Question

$\int \frac{e^{x}dx}{\sqrt{e^{2x}-1}}$ is equal to

• A. ${\sin }^{-1}{e}^x+C$
• B. $\ln (e^x+\sqrt{e^{2x}-1})+C$
• C. $\ln (e^x+\sqrt{e^{2x}+1})+C$
• D. ${\tan }^{-1}{e}^x+C$

Answer 1

The correct option is B $\ln (e^x+\sqrt{e^{2x}-1})+C$

we have,
$\int \frac{e^{x}dx}{\sqrt{e^{2x}-1}}$
Let, $e^x=t$
$\Rightarrow e^{x}dx=dt$
$\Rightarrow \int \frac{dt}{\sqrt{t^2-1}}=\ln (t+\sqrt{t^2-1})+C$
$=\ln (e^x+\sqrt{e^{2x}-1})+C$