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