Answer
$${x^2}{e^x} + {e^x} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\left( {x + 1} \right)}^2}{e^x}} dx \cr
& {\text{Use the integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}u = {\left( {x + 1} \right)^2},\,\,\,\,du = 2\left( {x + 1} \right)dx\,\,\,\,\, \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^x}dx,\,\,\,\,v = {e^x} \cr
& \cr
& {\text{Integration by parts then gives}} \cr
& \int {{{\left( {x + 1} \right)}^2}{e^x}} dx = {e^x}{\left( {x + 1} \right)^2} - \int {{e^x}\left( {2\left( {x + 1} \right)dx\,\,} \right)} \cr
& \int {{{\left( {x + 1} \right)}^2}{e^x}} dx = {e^x}{\left( {x + 1} \right)^2} - \int {2{e^x}\left( {x + 1} \right)dx\,\,} \cr
& \cr
& {\text{Integrate by parts again to get}}\int {2{e^x}\left( {x + 1} \right)dx\,\,} \cr
& \,\,\,{\text{Let }}u = 2\left( {x + 1} \right),\,\,\,\,du = 2dx\,\,\,\,\, \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^x}dx,\,\,\,\,v = {e^x} \cr
& \int {{{\left( {x + 1} \right)}^2}{e^x}} dx = {e^x}{\left( {x + 1} \right)^2} - \left( {2{e^x}\left( {x + 1} \right) - \int {{e^x}\left( {2dx} \right)} } \right) \cr
& \int {{{\left( {x + 1} \right)}^2}{e^x}} dx = {e^x}{\left( {x + 1} \right)^2} - 2{e^x}\left( {x + 1} \right) + 2\int {{e^x}dx} \cr
& \int {{{\left( {x + 1} \right)}^2}{e^x}} dx = {e^x}{\left( {x + 1} \right)^2} - 2{e^x}\left( {x + 1} \right) + 2{e^x} + C \cr
& \cr
& {\text{Expand and simplify}} \cr
& \int {{{\left( {x + 1} \right)}^2}{e^x}} dx = {e^x}\left( {{x^2} + 2x + 1} \right) - 2x{e^x} - 2{e^x} + 2{e^x} + C \cr
& \int {{{\left( {x + 1} \right)}^2}{e^x}} dx = {x^2}{e^x} + 2x{e^x} + {e^x} - 2x{e^x} - 2{e^x} + 2{e^x} + C \cr
& \int {{{\left( {x + 1} \right)}^2}{e^x}} dx = {x^2}{e^x} + {e^x} + C \cr} $$
