Answer
$$\frac{1}{{27}}{e^{3x}}\left( {9{x^2} - 6x + 2} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}{e^{3x}}} dx \cr
& = \frac{1}{3}\int {{x^2}{e^{3x}}\left( 3 \right)} dx \cr
& \left( {\text{a}} \right){\text{ Integrate by tables }} \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {{u^n}{e^u}du} = {u^n}{e^u} - n\int {{u^{n - 1}}{e^u}} du \cr
& \frac{1}{3}\int {{x^2}{e^{3x}}\left( 3 \right)} dx = \frac{1}{3}{x^2}{e^{3x}} - \frac{2}{3}\int {x{e^{3x}}} dx \cr
& = \frac{1}{3}{x^2}{e^{3x}} - \frac{2}{{27}}\int {3x{e^{3x}}\left( 3 \right)} dx \cr
& {\text{Use the formula }}\int {u{e^u}du} = \left( {u - 1} \right){e^u} + C \cr
& = \frac{1}{3}{x^2}{e^{3x}} - \frac{2}{{27}}\left( {3x - 1} \right){e^{3x}} + C \cr
& {\text{Simplifying}} \cr
& = \frac{1}{3}{x^2}{e^{3x}} - \frac{6}{{27}}x{e^{3x}} + \frac{2}{{27}}{e^{3x}} + C \cr
& {\text{Factoring}} \cr
& = \frac{1}{{27}}{e^{3x}}\left( {9{x^2} - 6x + 2} \right) + C \cr
& \cr
& \left( {\text{b}} \right){\text{Integrate by parts}} \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx \cr
& dv = {e^{3x}},{\text{ }}v = \frac{1}{3}{e^{3x}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {{x^2}{e^{3x}}} dx = \frac{1}{3}{x^2}{e^{3x}} - \int {\left( {\frac{1}{3}{e^{3x}}} \right)} \left( {2x} \right)dx \cr
& = \frac{1}{3}{x^2}{e^{3x}} - \frac{2}{3}\int {x{e^{3x}}} dx \cr
& {\text{Let }}u = x,{\text{ }}du = dx \cr
& dv = {e^{3x}},{\text{ }}v = \frac{1}{3}{e^{3x}} \cr
& = \frac{1}{3}{x^2}{e^{3x}} - \frac{2}{3}\left( {\frac{1}{3}x{e^{3x}} - \int {\left( {\frac{1}{3}{e^{3x}}} \right)dx} } \right) \cr
& = \frac{1}{3}{x^2}{e^{3x}} - \frac{2}{9}x{e^{3x}} + \frac{2}{9}\int {{e^{3x}}dx} \cr
& = \frac{1}{3}{x^2}{e^{3x}} - \frac{2}{9}x{e^{3x}} + \frac{2}{{27}}{e^{3x}} + C \cr
& {\text{Factoring}} \cr
& = \frac{1}{{27}}{e^{3x}}\left( {9{x^2} - 6x + 2} \right) + C \cr} $$
