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