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