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