Answer
$$\frac{{{x^2}{e^{3x}}}}{3} - \frac{{2x{e^{3x}}}}{9} + \frac{2}{{27}}{e^{3x}} + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}{e^{3x}}} dx \cr
& {\text{use the reduction formula }} \cr
& \int {{x^n}{e^{ax}}dx} = \frac{{{x^n}{e^{ax}}}}{a} - \frac{n}{a}\int {{x^{n - 1}}{e^{ax}}dx} \cr
& {\text{with }}n = 2,{\text{ }}a = 3 \cr
& \int {{x^2}{e^{3x}}} dx = \frac{{{x^2}{e^{3x}}}}{3} - \frac{2}{3}\int {{x^{2 - 1}}{e^{3x}}dx} \cr
& \int {{x^2}{e^{3x}}} dx = \frac{{{x^2}{e^{3x}}}}{3} - \frac{2}{3}\int {x{e^{3x}}dx} \cr
& {\text{use the reduction formula with }}n = 1,{\text{ }}a = 3 \cr
& \int {{x^2}{e^{3x}}} dx = \frac{{{x^2}{e^{3x}}}}{3} - \frac{2}{3}\left( {\frac{{x{e^{3x}}}}{3} - \frac{1}{3}\int {{x^{1 - 1}}{e^{3x}}dx} } \right) \cr
& \int {{x^2}{e^{3x}}} dx = \frac{{{x^2}{e^{3x}}}}{3} - \frac{2}{3}\left( {\frac{{x{e^{3x}}}}{3} - \frac{1}{3}\int {{e^{3x}}dx} } \right) \cr
& \int {{x^2}{e^{3x}}} dx = \frac{{{x^2}{e^{3x}}}}{3} - \frac{{2x{e^{3x}}}}{9} + \frac{2}{9}\int {{e^{3x}}dx} \cr
& {\text{integrating}} \cr
& \int {{x^2}{e^{3x}}} dx = \frac{{{x^2}{e^{3x}}}}{3} - \frac{{2x{e^{3x}}}}{9} + \frac{2}{9}\left( {\frac{1}{3}{e^{3x}}} \right) + C \cr
& \int {{x^2}{e^{3x}}} dx = \frac{{{x^2}{e^{3x}}}}{3} - \frac{{2x{e^{3x}}}}{9} + \frac{2}{{27}}{e^{3x}} + C \cr} $$
