Answer
$$\frac{{{x^2}{e^{2x}}}}{2} - \frac{{x{e^{2x}}}}{2} + \frac{{{e^{2x}}}}{4} + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}} {e^{2x}}dx \cr
& {\text{setting }}\,\,\,\,\,\,u = {x^2}{\text{ then }}du = 2xdx\,\,\,\,\,\,\,\,\,\, \cr
& {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^{2x}}dx{\text{ then }}v = \frac{1}{2}{e^{2x}} \cr
& {\text{Substituting these values into the formula for integration by parts}} \cr
& \int u dv = uv - \int {vdu} \cr
& \int {{x^2}} {e^{2x}}dx = \frac{{{x^2}{e^{2x}}}}{2} - \int {\left( {\frac{1}{2}{e^{2x}}} \right)} \left( {2xdx} \right) \cr
& \int {{x^2}} {e^{2x}}dx = \frac{{{x^2}{e^{2x}}}}{2} - \int {x{e^{2x}}} dx \cr
& \cr
& {\text{integrate by parts again }}\int {x{e^{2x}}} dx \cr
& {\text{setting }}\,\,\,\,\,\,u = x{\text{ then }}du = dx\,\,\,\,\,\,\,\,\,\, \cr
& {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^{2x}}dx{\text{ then }}v = \frac{1}{2}{e^{2x}} \cr
& {\text{then}} \cr
& \int {{x^2}} {e^{2x}}dx = \frac{{{x^2}{e^{2x}}}}{2} - \left( {\frac{x}{2}{e^{2x}} - \int {\frac{1}{2}{e^{2x}}dx} } \right) \cr
& \int {{x^2}} {e^{2x}}dx = \frac{{{x^2}{e^{2x}}}}{2} - \frac{x}{2}{e^{2x}} + \int {\frac{1}{2}{e^{2x}}dx} \cr
& {\text{integrate}} \cr
& \int {{x^2}} {e^{2x}}dx = \frac{{{x^2}{e^{2x}}}}{2} - \frac{{x{e^{2x}}}}{2} + \frac{1}{2}\left( {\frac{{{e^{2x}}}}{2}} \right) + C \cr
& \int {{x^2}} {e^{2x}}dx = \frac{{{x^2}{e^{2x}}}}{2} - \frac{{x{e^{2x}}}}{2} + \frac{{{e^{2x}}}}{4} + C \cr} $$
