Answer
$${x^2}{e^x} - 2x{e^x} + 4{e^x} + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {{x^2} + 2} \right){e^x}} dx \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = {x^2} + 2,{\text{ }}du = 2xdx{\text{ }} \cr
& dv = {e^x}dx,{\text{ }}v = {e^x} \cr
& {\text{Using the formula of integration by parts}} \cr
& \int u dv = uv - \int v du \cr
& \int {\left( {{x^2} + 2} \right){e^x}} dx = \left( {{x^2} + 2} \right){e^x} - \int {\left( {{e^x}} \right)} \left( {2x} \right)dx \cr
& \int {\left( {{x^2} + 2} \right){e^x}} dx = \left( {{x^2} + 2} \right){e^x} - \int {2x{e^x}} dx{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Integrate by parts }}\int {2x{e^x}} dx{\text{,}} \cr
& {\text{Let }}u = 2x,{\text{ }}du = 2dx{\text{ }} \cr
& dv = {e^x}dx,{\text{ }}v = {e^x} \cr
& \int u dv = uv - \int v du \cr
& \int {2x{e^x}} dx = 2x{e^x} - 2\int {{e^x}} dx \cr
& = 2x{e^x} - 2{e^x} + C \cr
& {\text{Substituting the previous result into }}\left( {\bf{1}} \right) \cr
& \int {\left( {{x^2} + 2} \right){e^x}} dx = \left( {{x^2} + 2} \right){e^x} - \left( {2x{e^x} - 2{e^x}} \right) + C \cr
& = \left( {{x^2} + 2} \right){e^x} - 2x{e^x} + 2{e^x} + C \cr
& = {x^2}{e^x} + 2{e^x} - 2x{e^x} + 2{e^x} + C \cr
& = {x^2}{e^x} - 2x{e^x} + 4{e^x} + C \cr} $$
