Question

$\int e^{2x+\log x}dx=$

• A. $\frac{1}{4}(2x-1)e^{2x}+c$
• B. $\frac{1}{4}(2x+1)e^{2x}+c$
• C. $\frac{1}{2}(2x+1)e^{2x}+c$
• D. $\frac{1}{2}(2x-1)e^{2x}+c$

Answer 1

Answer: (A)

$\int e^{2x+\log x}dx=\int x{e}^{2x}dx$
$=\frac{x{e}^{2x}}{2}-\int \frac{1}{2}{e}^{2x}dx+c=\frac{e^{2x}}{4}(2x-1)+c$