Answer
$\frac{1}{3}{x^2}{e^{3x}} - \frac{2}{9}x{e^{3x}} + \frac{{11}}{{27}}{e^{3x}} + C$
Work Step by Step
$$\eqalign{
& \int {\left( {1 + {x^2}} \right)} {e^{3x}}dx \cr
& \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = 1 + {x^2},{\text{ }}du = 2xdx \cr
& dv = {e^{3x}}dx,{\text{ }}v = \frac{1}{3}{e^{3x}} \cr
& {\text{Using the integration by parts formula}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {\left( {1 + {x^2}} \right)} {e^{3x}}dx = \left( {1 + {x^2}} \right)\left( {\frac{1}{3}{e^{3x}}} \right) - \int {\left( {\frac{1}{3}{e^{3x}}} \right)} \left( {2x} \right)dx \cr
& \int {\left( {1 + {x^2}} \right)} {e^{3x}}dx = \frac{1}{3}\left( {1 + {x^2}} \right){e^{3x}} - \frac{2}{3}\int {x{e^{3x}}} dx \cr
& \cr
& {\text{Integate by parts again}} \cr
& {\text{Let }}u = x,{\text{ }}du = dx \cr
& dv = {e^{3x}}dx,{\text{ }}v = \frac{1}{3}{e^{3x}} \cr
& \int {\left( {1 + {x^2}} \right)} {e^{3x}}dx = \frac{1}{3}\left( {1 + {x^2}} \right){e^{3x}} - \frac{2}{3}\left( {x\left( {\frac{1}{3}{e^{3x}}} \right) - \int {\frac{1}{3}{e^{3x}}dx} } \right) \cr
& {\text{Multiply}} \cr
& \int {\left( {1 + {x^2}} \right)} {e^{3x}}dx = \frac{1}{3}{e^{3x}} + \frac{1}{3}{x^2}{e^{3x}} - \frac{2}{9}x{e^{3x}} + \frac{2}{9}\int {{e^{3x}}dx} \cr
& {\text{Integrate and simplify}} \cr
& \int {\left( {1 + {x^2}} \right)} {e^{3x}}dx = \frac{1}{3}{e^{3x}} + \frac{1}{3}{x^2}{e^{3x}} - \frac{2}{9}x{e^{3x}} + \frac{2}{9}\left( {\frac{1}{3}{e^{3x}}} \right) + C \cr
& \int {\left( {1 + {x^2}} \right)} {e^{3x}}dx = \frac{1}{3}{x^2}{e^{3x}} + \frac{1}{3}{e^{3x}} - \frac{2}{9}x{e^{3x}} + \frac{2}{{27}}{e^{3x}} + C \cr
& \int {\left( {1 + {x^2}} \right)} {e^{3x}}dx = \frac{1}{3}{x^2}{e^{3x}} - \frac{2}{9}x{e^{3x}} + \frac{{11}}{{27}}{e^{3x}} + C \cr} $$
