Answer
$${e^{{x^2} + x}} - \frac{{{x^2}{e^{2x + 1}}}}{2} + \frac{x}{2}{e^{2x + 1}} - \frac{1}{4}{e^{2x + 1}} + C$$
Work Step by Step
$$\eqalign{
& \int {\left[ {\left( {2x + 1} \right){e^{{x^2} + x}} - {x^2}{e^{2x + 1}}} \right]} dx \cr
& {\text{Distribute}} \cr
& \int {\left( {2x + 1} \right){e^{{x^2} + x}}} dx - \int {{x^2}{e^{2x + 1}}} dx{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{*The first integral can be solved by substitution}} \cr
& {\text{*The second integral can be solved by parts}} \cr
& \to {\text{ }}\int {\left( {2x + 1} \right){e^{{x^2} + x}}} dx \cr
& {\text{ }}u = {x^2} + x,{\text{ }}du = \left( {2x + 1} \right)dx \cr
& \int {\left( {2x + 1} \right){e^{{x^2} + x}}} dx = \int {{e^u}} du = {e^u} + C = {e^{{x^2} + x}} + C \cr
& \cr
& \to \int {{x^2}{e^{2x + 1}}} dx \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx \cr
& dv = {e^{2x + 1}}dx,{\text{ }}v = \frac{1}{2}{e^{2x + 1}} \cr
& \int u dv = uv - \int v du \cr
& \int {{x^2}{e^{2x + 1}}} dx = {x^2}\left( {\frac{1}{2}{e^{2x + 1}}} \right) - \int {\left( {\frac{1}{2}{e^{2x + 1}}} \right)} \left( {2x} \right)dx \cr
& \int {{x^2}{e^{2x + 1}}} dx = \frac{{{x^2}{e^{2x + 1}}}}{2} - \int {x{e^{2x + 1}}} dx \cr
& {\text{Integrate by parts }}\int {x{e^{2x + 1}}} dx{\text{,}} \cr
& {\text{Let }}u = x,{\text{ }}du = dx \cr
& dv = {e^{2x + 1}}dx,{\text{ }}v = \frac{1}{2}{e^{2x + 1}} \cr
& {\text{ }}\int {x{e^{2x + 1}}} dx{\text{,}} = \left( x \right)\left( {\frac{1}{2}{e^{2x + 1}}} \right) - \int {\frac{1}{2}{e^{2x + 1}}dx} \cr
& {\text{ }}\int {x{e^{2x + 1}}} dx{\text{,}} = \frac{1}{2}x{e^{2x + 1}} - \frac{1}{4}{e^{2x + 1}} + C \cr
& {\text{Then,}} \cr
& \int {{x^2}{e^{2x + 1}}} dx = \frac{{{x^2}{e^{2x + 1}}}}{2} - \frac{1}{2}x{e^{2x + 1}} + \frac{1}{4}{e^{2x + 1}} + C \cr
& {\text{Substituting both results into }}\left( {\bf{1}} \right) \cr
& \int {\left( {2x + 1} \right){e^{{x^2} + x}}} dx - \int {{x^2}{e^{2x + 1}}} dx{\text{ }}\left( {\bf{1}} \right) \cr
& = {e^{{x^2} + x}} - \frac{{{x^2}{e^{2x + 1}}}}{2} + \frac{x}{2}{e^{2x + 1}} - \frac{1}{4}{e^{2x + 1}} + C \cr} $$
