## Calculus, 10th Edition (Anton)

$$\frac{{{x^2}}}{2}{e^{ - 2x}} - \frac{x}{2}{e^{ - 2x}} + \frac{1}{4}{e^{ - 2x}} + C$$
\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}