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) \[\frac{1}{2}(2x+1){{e}^{2x}}+c\]
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{{{e}^{2x+\log x}}dx}=\int_{{}}^{{}}{x{{e}^{2x}}dx}\] \[=\frac{x{{e}^{2x}}}{2}-\int_{{}}^{{}}{\frac{1}{2}{{e}^{2x}}dx+c}=\frac{{{e}^{2x}}}{4}(2x-1)+c.\]
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