Answer
$\frac{1}{8}{e^{2x}}\left( {4{x^3} - 6{x^2} + 6x - 3} \right) + C$
Work Step by Step
$$\eqalign{
& \int {{x^3}{e^{2x}}} dx \cr
& \int {{u^n}{e^{au}}du} = \frac{1}{a}{u^n}{e^{au}} - \frac{n}{a}\int {{u^{n - 1}}{e^{au}}du} \cr
& \cr
& {\text{Let }}u = x,{\text{ }}du = dx,{\text{ }}n = 3{\text{ and }}a = 2,{\text{ we obtain}} \cr
& \int {{x^3}{e^{2x}}} dx = \frac{1}{2}{x^3}{e^{2x}} - \frac{3}{2}\int {{x^{3 - 1}}{e^{2x}}dx} \cr
& = \frac{1}{2}{x^3}{e^{2x}} - \frac{3}{2}\int {{x^2}{e^{2x}}dx} \cr
& \cr
& {\text{For }}\int {{x^2}{e^{2x}}dx} {\text{ we have}}{\text{, }} \cr
& u = x,{\text{ }}du = dx,{\text{ }}n = 2{\text{ and }}a = 2 \cr
& = \frac{1}{2}{x^3}{e^{2x}} - \frac{3}{2}\left( {\frac{1}{2}{x^2}{e^{2x}} - \int {{x^{2 - 1}}{e^{2x}}dx} } \right) \cr
& = \frac{1}{2}{x^3}{e^{2x}} - \frac{3}{4}{x^2}{e^{2x}} + \frac{3}{2}\int {x{e^{2x}}dx} \cr
& {\text{Therefore: }}\int {u{e^{au}}du} = \frac{1}{{{a^2}}}\left( {au - 1} \right){e^{au}} + C,{\text{ }}a = 2 \cr
& = \frac{1}{2}{x^3}{e^{2x}} - \frac{3}{4}{x^2}{e^{2x}} + \frac{3}{{2{{\left( 2 \right)}^2}}}\left( {2x - 1} \right){e^{2x}} + C \cr
& = \frac{1}{2}{x^3}{e^{2x}} - \frac{3}{4}{x^2}{e^{2x}} + \frac{3}{8}\left( {2x - 1} \right){e^{2x}} + C \cr
& = \frac{1}{2}{x^3}{e^{2x}} - \frac{3}{4}{x^2}{e^{2x}} + \frac{3}{4}x{e^{2x}} - \frac{3}{8}{e^{2x}} + C \cr
& {\text{Factoring out }}\frac{1}{8}{e^{2x}} \cr
& = \frac{1}{8}{e^{2x}}\left( {4{x^3} - 6{x^2} + 6x - 3} \right) + C \cr} $$
