Answer
$$\frac{{{x^2}}}{4}{e^{4x}} - \frac{x}{8}{e^{4x}} + \frac{1}{{32}}{e^{4x}} + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}{e^{4x}}} dx \cr
& {\text{substitute }}u = {x^2},{\text{ }}du = 2xdx \cr
& dv = {e^{4x}},{\text{ }}v = \frac{1}{4}{e^{4x}} \cr
& {\text{applying integration by parts}}{\text{, we have}} \cr
& \int {{x^2}{e^{4x}}} dx = \frac{{{x^2}}}{4}{e^{4x}} - \int {\left( {\frac{1}{4}{e^{4x}}} \right)\left( {2xdx} \right)} \cr
& \int {{x^2}{e^{4x}}} dx = \frac{{{x^2}}}{4}{e^{4x}} - \frac{1}{2}\int {x{e^{4x}}dx} \cr
& {\text{substitute }}u = x,{\text{ }}du = dx \cr
& dv = {e^{4x}},{\text{ }}v = \frac{1}{4}{e^{4x}} \cr
& {\text{applying integration by parts}}{\text{, we have}} \cr
& \int {{x^2}{e^{4x}}} dx = \frac{{{x^2}}}{4}{e^{4x}} - \frac{1}{2}\left( {\frac{x}{4}{e^{4x}} - \int {\frac{1}{4}{e^{4x}}dx} } \right) \cr
& \int {{x^2}{e^{4x}}} dx = \frac{{{x^2}}}{4}{e^{4x}} - \frac{x}{8}{e^{4x}} + \frac{1}{8}\int {{e^{4x}}dx} \cr
& {\text{integrating}} \cr
& \int {{x^2}{e^{4x}}} dx = \frac{{{x^2}}}{4}{e^{4x}} - \frac{x}{8}{e^{4x}} + \frac{1}{8}\left( {\frac{1}{4}{e^{4x}}} \right) + C \cr
& \int {{x^2}{e^{4x}}} dx = \frac{{{x^2}}}{4}{e^{4x}} - \frac{x}{8}{e^{4x}} + \frac{1}{{32}}{e^{4x}} + C \cr} $$
