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