Integral Calculus Question 275
Question: $ \int_{{}}^{{}}{{e^{2x+\log x}}}dx= $
Options:
A) $ \frac{1}{4}(2x-1)+\frac{2}{x+1}+c $
B) $ \frac{1}{4}(2x+1)+\frac{2}{x+1}+c $
C) $ \frac{1}{2}(2x+1)e^{2x}+c $
D) $ \frac{1}{2}(2x+1)e^{2x}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{{e^{2x+\log x}}dx}=\int_{{}}^{{}}{xe^{2x}dx} $ $ =\frac{xe^{2x}}{2}-\int_{{}}^{{}}{\frac{1}{2}e^{2x}dx+c}=\frac{e^{2x}}{4}(2x-1)+c. $
BETA
