Answer
$$-\frac{1}{2}x{e^{ - 2x}} - \frac{1}{4}{e^{ - 2x}} + C$$
Work Step by Step
$$\eqalign{
& \int {x{e^{ - 2x}}} dx \cr
& {\text{substitute }}u = x,{\text{ }}du = dx \cr
& dv = {e^{ - 2x}}dx,{\text{ }}v = - \frac{1}{2}{e^{ - 2x}} \cr
& {\text{ integration by parts}} \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{, we have}} \cr
& \int {x{e^{ - 2x}}} dx = - \frac{1}{2}x{e^{ - 2x}} - \int {\left( { - \frac{1}{2}{e^{ - 2x}}} \right)dx} \cr
& \int {x{e^{ - 2x}}} dx = - \frac{1}{2}x{e^{ - 2x}} + \frac{1}{2}\int {{e^{ - 2x}}dx} \cr
& {\text{find antiderivative}} \cr
& \int {x{e^{ - 2x}}} dx = - \frac{1}{2}x{e^{ - 2x}} + \frac{1}{2}\left( { - \frac{1}{2}{e^{ - 2x}}} \right) + C \cr
& \int {x{e^{ - 2x}}} dx = - \frac{1}{2}x{e^{ - 2x}} - \frac{1}{4}{e^{ - 2x}} + C \cr} $$
