Integral Calculus Question 366
Question: $ \int_{{}}^{{}}{\frac{e^{x}(x-1)}{x^{2}}\ dx=} $
Options:
A) $ \frac{1}{x}e^{x}+c $
B) $ x{e^{-x}}+c $
C) $ \frac{1}{x^{2}}e^{x}+c $
D) $ ( x-\frac{1}{x} )e^{x}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{e^{x}(x-1)}{x^{2}},dx}=\int_{{}}^{{}}{e^{x}[ \frac{1}{x}-\frac{1}{x^{2}} ]},dx=\frac{e^{x}}{x}+c. $
BETA
