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