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Question Number 181750 by Mastermind last updated on 29/Nov/22

$$\int {\mathrm{xe}}^{{\mathrm{x}}^2+\mathrm{x}}\mathrm{dx}$$$$$$$$.$$

Commented by CElcedricjunior last updated on 29/Nov/22

$$\int {\mathbf{xe}}^{{\mathbf{x}}^2+\mathbf{x}}\mathbf{dx}=\frac{1}{2}\int (2\mathit{x}+1){\mathit{e}}^{{\mathit{x}}^2+\mathit{x}}\mathit{d}\mathit{x}-\frac{1}{2}\int {\mathit{e}}^{\mathit{x}+{\mathit{x}}^2}\mathit{d}\mathit{x}$$$$=\frac{1}{2}{\mathit{e}}^{\mathit{x}+{\mathit{x}}^2}-\frac{1}{2}\int {\mathit{e}}^{\mathit{x}+{\mathit{x}}^2}\mathit{d}\mathit{x}$$$$\mathit{p}\mathit{o}\mathit{s}\mathit{o}\mathit{n}\mathit{s}\{\begin{array}{l}\mathit{u}={\mathit{e}}^{\mathit{x}+{\mathit{x}}^2}\\ {\mathit{v}}^{\prime }=1\end{array}=\{\begin{array}{l}{\mathit{u}}^{\prime }=(2x+1)e^{\mathit{x}+{\mathit{x}}^2}\\ \mathit{v}=\mathit{x}\end{array}$$$$=\frac{1}{2}{\mathit{e}}^{\mathit{x}+{\mathit{x}}^2}-\frac{1}{2}{\mathit{x}\mathit{e}}^{\mathit{x}+{\mathit{x}}^2}+\frac{1}{2}\int \mathit{x}(2\mathit{x}+1){\mathit{e}}^{\mathit{x}+{\mathit{x}}^2}$$

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