Integral Calculus Question 327
Question: $ \int_{{}}^{{}}{\frac{1}{{\log_{x}}e}dx=} $
[MP PET 1994]
Options:
A) $ \log {\log_{x}}e+c $
B) $ \frac{1}{{{({\log_{x}}e)}^{2}}}+c $
C) $ x\log ( \frac{x}{e} )+c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{1}{{\log_{x}}e},dx=\int_{{}}^{{}}{{\log_{e}}x,dx}=x\log x-x+c} $ $ =x({\log_{e}}x-{\log_{e}}e)+c=x{\log_{e}}( \frac{x}{e} )+c. $
BETA
