Q. The value of the integral $\int_{}^{}\frac{dx}{\left( e^{x} + e^{- x} \right)^{2}}$ is
A) $\frac{1}{2}\left( e^{2x} + 1 \right) + c$
B) $\frac{1}{2}\left( e^{- 2x} + 1 \right) + c$
C) $- \frac{1}{2}\left( e^{2x} + 1 \right)^{- 1} + c$
D) $\frac{1}{4}\left( e^{2x} - 1 \right) + c$
Explanation
(c)
Let $I = \int_{}^{}\frac{dx}{\left( e^{x} + e^{- x} \right)^{2}} = \int_{}^{}\frac{e^{2x}\ \ dx}{\left( e^{2x} + 1 \right)^{2}}$
Put $e^{2x} + 1 = z$
$\Rightarrow \ 2e^{2x}dx = dz$
$\therefore\ I = \frac{1}{2}\ \int_{}^{}\frac{dz}{z^{2}}$
$= - \frac{1}{2z} + c$
$= - \frac{1}{2(e^{2x} + 1)} + c$
$= - \frac{1}{2}\left( e^{2x} + 1 \right)^{- 1} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.