# Chapter 8: Techniques of Integration - Practice Exercises - Page 517: 5

$${x^2}{e^x} + {e^x} + C$$

#### 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}

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