Q. $\int_{}^{}(e^{x} + e^{- x}).\left( e^{x} - e^{- x} \right)dx$ is equal to
A) $e^{x} + c$
B) $\frac{1}{2}\ \left( e^{x} - e^{- x} \right)^{2} + c$
C) $\frac{1}{2}\left( e^{x} + e^{- x} \right)^{2} + c$
D) $\frac{1}{3}\left( e^{x} + e^{- x} \right)^{3} + c$
Explanation
(d)
Let $I = \int_{}^{}\left( e^{x} + e^{- x} \right)^{2}.(e^{x} - e^{- x})dx$
Put $e^{x} + e^{- x} = t\ \ \Rightarrow \ \ \ \left( e^{x} - e^{- x} \right)dx = dt$
$\therefore\ \ I = \int_{}^{}{t^{2}dt = \frac{t^{3}}{3}} + c\$
$= \frac{\left( e^{x} + e^{- x} \right)^{3}}{3} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.