Question

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

• A. $x-\log [1+\sqrt{1-e^{2x}}]+C$
• B. $\log (e^x+1)-x-e^{-x}+C$
• C. $\log (e^{2x}+1)-x+c$
• D. $\frac{1}{2(e^{2x}+1)}+c$

Answer 1

The correct option is A $x-\log [1+\sqrt{1-e^{2x}}]+C$

$I=\int \frac{1}{\sqrt{1-e^{2x}}}dx=\int \frac{e^{-x}}{\sqrt{e^{-2x}-1}}dx$
Put $e^{-x}=t\Rightarrow -e^{-x}dx=dt,$
$\Rightarrow I=-\int \frac{1}{\sqrt{t^2-1}}dt$
$=-\log [t+\sqrt{t^2-1}]+C$
$-\log [e^{-x}+\sqrt{e^{-2x}-1}]+C$
$=-\log [\frac{1}{e^x}+\frac{\sqrt{1-e^{2x}}}{e^x}]+C$
$=-\log [1+\sqrt{1-e^{2x}}]+\log e^x+C$
$=x-\log [1+\sqrt{1-e^{2x}}]+C$