Question

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

• A. $-\log |e^{-x}+\sqrt{e^{-2x}-1}|+C$
• B. $-\log |e^x+\sqrt{e^{-2x}-1}|+C$
• C. $-\log |e^x+\sqrt{e^{-x}-1}|+C$
• D. $-\log |e^{-x}+\sqrt{e^{-2x}+1}|+C$

Answer 1

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

$I=\int \frac{1}{\sqrt{1-e^{2x}}}dx=\int \frac{1}{\sqrt{1-\frac{1}{e^{-2x}}}}dx=\int \frac{e^{-x}}{\sqrt{e^{-2x}-1}}dx$

$=\int \frac{e^{-x}}{\sqrt{{\left(e^{-x}\right)}^2-1^2}}dx$

Let $e^{-x}=t$ $\Rightarrow -e^{-x}dx=dt$

$\therefore I=-\int \frac{dt}{\sqrt{t^2-1^2}}=-\log |t+\sqrt{t^2-1}|+C$

$=-\log |e^{-x}+\sqrt{e^{-2x}-1}|+C$