Question

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

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

Answer 1

The correct option is C

$-\log \left|e^{-x}+\sqrt{e^{-2x}–1}\right|+c$

Explanation for the correct option:

Evaluate the integral:

$\begin{array}{l}\begin{array}{c}\int \frac{dx}{\sqrt{1-e^{2x}}}=\int \frac{e^{-x}dx}{\sqrt{e^{-2x}(1-e^{2x})}}\\ =\int \frac{e^{-x}dx}{\sqrt{e^{-2x}-1}}\\ \text{Let}{e}^{-x}=z\\ -e^{-x}dx=dz\\ \begin{array}{c}=-\int \frac{dz}{\sqrt{z^2-1}}\\ =-\log \left|z+\sqrt{z^2-1}\right|+c\\ =-\log \left|e^{-x}+\sqrt{e^{-2x}-1}\right|+c\end{array}\end{array}\end{array}$

Therefore, $\int \frac{dx}{\sqrt{1-e^{2x}}}=$$-\log \left|e^{-x}+\sqrt{e^{-2x}–1}\right|+c$

Hence the correct answer is option (C).