Answer
$$\frac{{{x^3}{e^{2x}}}}{2} - \frac{{3{x^2}{e^{2x}}}}{4} + \frac{{3x{e^x}}}{4} - \frac{3}{8}{e^{2x}} + C$$
Work Step by Step
$$\eqalign{
& \int {{x^3}{e^{2x}}} dx \cr
& {\text{From formula 70 on this page}} \cr
& \int {{x^n}{e^{ax}}dx} = \frac{{{x^n}{e^{ax}}}}{a} - \frac{n}{a}\int {{x^{n - 1}}{e^{ax}}dx} \cr
& {\text{Let }}n = 3,{\text{ }}a = 2 \cr
& \int {{x^3}{e^{2x}}} dx = \frac{{{x^3}{e^{2x}}}}{2} - \frac{3}{2}\int {{x^{3 - 1}}{e^{2x}}dx} \cr
& \int {{x^3}{e^{2x}}} dx = \frac{{{x^3}{e^{2x}}}}{2} - \frac{3}{2}\int {{x^2}{e^{2x}}dx} \cr
& {\text{To solve }}\int {{x^2}{e^{2x}}dx} {\text{, let }}n = 2,{\text{ }}a = 2 \cr
& \int {{x^3}{e^{2x}}} dx = \frac{{{x^3}{e^{2x}}}}{2} - \frac{3}{2}\left( {\frac{{{x^2}{e^{2x}}}}{2} - \frac{2}{2}\int {{x^{2 - 1}}{e^{2x}}dx} } \right) \cr
& \int {{x^3}{e^{2x}}} dx = \frac{{{x^3}{e^{2x}}}}{2} - \frac{{3{x^2}{e^{2x}}}}{4} + \frac{3}{2}\int {x{e^{2x}}dx} \cr
& {\text{To solve }}\int {x{e^{2x}}dx} {\text{, let }}n = 1,{\text{ }}a = 2 \cr
& \int {{x^3}{e^{2x}}} dx = \frac{{{x^3}{e^{2x}}}}{2} - \frac{{3{x^2}{e^{2x}}}}{4} + \frac{3}{2}\left( {\frac{{x{e^x}}}{2} - \frac{1}{2}\int {{e^{2x}}dx} } \right) \cr
& \int {{x^3}{e^{2x}}} dx = \frac{{{x^3}{e^{2x}}}}{2} - \frac{{3{x^2}{e^{2x}}}}{4} + \frac{{3x{e^x}}}{4} - \frac{3}{4}\int {{e^{2x}}dx} \cr
& \int {{x^3}{e^{2x}}} dx = \frac{{{x^3}{e^{2x}}}}{2} - \frac{{3{x^2}{e^{2x}}}}{4} + \frac{{3x{e^x}}}{4} - \frac{3}{4}\left( {\frac{1}{2}{e^{2x}}} \right) + C \cr
& \int {{x^3}{e^{2x}}} dx = \frac{{{x^3}{e^{2x}}}}{2} - \frac{{3{x^2}{e^{2x}}}}{4} + \frac{{3x{e^x}}}{4} - \frac{3}{8}{e^{2x}} + C \cr} $$
