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)

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