Q. $\int_{}^{}\frac{dx}{\sqrt{1 - e^{2x}}}$ is equal to
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) $- \log{\left| e^{- 2x} + \sqrt{e^{- 2x} - 1} \right| + c}$
Explanation
(c)
Let $I = \int_{}^{}\frac{dx}{\sqrt{1 - e^{2x}}} = \int_{}^{}\frac{e^{- x}}{\sqrt{e^{- 2x} - 1}}\ dx$
Put $e^{- x} = t \Rightarrow e^{- x}dx = - dt$
$\therefore\ I = - \int_{}^{}\frac{dt}{\sqrt{t^{2} - 1}} = - \log{\left| t + \sqrt{t^{2} - 1} \right| + c}$
$= - \log{\left| e^{- x} + \sqrt{e^{- 2x} - 1} \right| + c}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.