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Question Number 98594 by bemath last updated on 15/Jun/20

Answered by bobhans last updated on 15/Jun/20

$$\int {\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}({\mathrm{x}}^2+2\mathrm{x}+1)\mathrm{dx}=\int {\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}{\mathrm{x}}^2\mathrm{dx}+\int 2\mathrm{x}{\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}\mathrm{dx}+\int {\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}\mathrm{dx}$$$${\mathrm{I}}_1=\int {\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}{\mathrm{x}}^2\mathrm{dx}=\int \mathrm{x}{\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}\mathrm{d}\left(\frac{{\mathrm{x}}^2}{2}\right)={\mathrm{xe}}^{\frac{{\mathrm{x}}^2}{2}}-\int {\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}\mathrm{dx}$$$${\mathrm{I}}_2=\int {2\mathrm{xe}}^{\frac{{\mathrm{x}}^2}{2}}\mathrm{dx}=2\int {\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}\mathrm{d}\left(\frac{{\mathrm{x}}^2}{2}\right)={2\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}$$$${\mathrm{I}}_3=\int {\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}\mathrm{dx}$$$$\therefore \mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3={\mathrm{xe}}^{\frac{{\mathrm{x}}^2}{2}}+{2\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}=(\mathrm{x}+2){\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}+\mathrm{C}$$

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