Question

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

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

Answer 1

Answer: (C)

Let $I=\int \frac{dx}{\sqrt{1-e^{2x}}}=\int \frac{e^{-x}}{\sqrt{e^{-2x}-1}}dx$ Put $t=e^{-x}$
$\Rightarrow$ $dt=-e^{-x}dx$
$\therefore$ $I=\int -\frac{dt}{\sqrt{t^2-1}}=-\log |t+\sqrt{t^2-1}|+c$
$=-\log |e^{-x}+\sqrt{e^{-2x}-1}+c$