Answer
$= \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} + C$
Work Step by Step
$\int xe^{2x}$
$u = x$
$u' = 1$
$\frac{du}{dx} = 1$
$du = 1 dx$
$du = dx$
$dv = e^{2x}$
$v = \frac{1}{2}e^{2x}$
$uv - \int vdu$
$= (x)(\frac{1}{2}e^{2x}) - \int \frac{1}{2}e^{2x}dx$
$= (x)(\frac{1}{2}e^{2x}) - \frac{1}{2}\times\frac{1}{2}e^{2x}$
$= (x)(\frac{1}{2}e^{2x}) - \frac{1}{4}e^{2x}$
$= \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x}$
$= \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} + C$