Definite Integration Question 150
Question: $ \int_1^{e}{\frac{e^{x}}{x}(1+x\log x)dx}= $
Options:
A) $ e^{e} $
B) $ e^{e}-e $
C) $ e^{e}+e $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_1^{e}{\frac{e^{x}}{x}(1+x\log x)dx=\int_1^{e}{\frac{1}{x}e^{x}dx}} $
$ +\int_1^{e}{e^{x}{\log _{e}}xdx} $
= $ [e^{x}\log x]_1^{e}-\int_1^{e}{e^{x}\log xdx+\int_1^{e}{e^{x}\log xdx}} $
= $ [e^{e}\log e-e^{1}{\log _{e}}1]=e^{e} $ .
BETA
