Integral Calculus Question 367
Question: $ \int_{{}}^{{}}{e^{x}[ \frac{1+x\log x}{x} ]\ dx=} $
Options:
A) $ e^{x}+\log x+c $
B) $ \frac{e^{x}}{\log x}+c $
C) $ e^{x}-\log x+c $
D) $ e^{x}\log x+c $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{e^{x}[ \frac{1+x\log x}{x} ],dx=}\int_{{}}^{{}}{e^{x}( \log x+\frac{1}{x} ),dx}=e^{x}\log x+c $ .
BETA
