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Question Number 95020 by 174 last updated on 22/May/20

Commented by EmericGent last updated on 22/May/20

There is no standard expression of this thing

Answered by niroj last updated on 22/May/20

$$\mathrm{I}=\int ({\mathrm{x}}^2-\mathrm{x}){\mathrm{e}}^{{\mathrm{x}}^2}\mathrm{dx}$$$$=\int {\mathrm{x}}^2{\mathrm{e}}^{{\mathrm{x}}^2}\mathrm{dx}-\int {\mathrm{xe}}^{{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{I}=I_1+I_2$$$$I_2=\int {\mathrm{xe}}^{{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{Put},{\mathrm{x}}^2=\mathrm{t}$$$$2\mathrm{xdx}=\mathrm{dt}$$$$\mathrm{xdx}=\frac{1}{2}\mathrm{dt}$$$${\mathrm{I}}_2=\int {\mathrm{e}}^{\mathrm{t}}\frac{1}{2}\mathrm{dt}=\frac{1}{2}{\mathrm{e}}^{\mathrm{t}}+\mathrm{C}=\frac{1}{2}{\mathrm{e}}^{{\mathrm{x}}^2}+\mathrm{C}$$$${\mathrm{I}}_1=\mathrm{x}\int \mathrm{x}.{\mathrm{e}}^{{\mathrm{x}}^2}\mathrm{dx}-\int (1.\int {\mathrm{xe}}^{{\mathrm{x}}^2}\mathrm{dx})\mathrm{dx}$$$$=\mathrm{x}.\frac{1}{2}{\mathrm{e}}^{{\mathrm{x}}^2}-\frac{1}{2}\int {\mathrm{e}}^{{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2$$$$=\frac{1}{2}{\mathrm{xe}}^{{\mathrm{x}}^2}-\frac{1}{2}\int {\mathrm{e}}^{{\mathrm{x}}^2}\mathrm{dx}+\frac{1}{2}{\mathrm{e}}^{{\mathrm{x}}^2}+\mathrm{C}//.$$$$=\frac{1}{2}{\mathrm{e}}^{{\mathrm{x}}^2}(\mathrm{x}+1)-\frac{1}{2}\left[\frac{\sqrt{\pi }}{2}\mathrm{error}\mathrm{f}\mathrm{i}\left(\mathrm{x}\right)\right]+\mathrm{C}$$$$N\mathrm{ow}\mathrm{use}\mathrm{more}\mathrm{complex}\mathrm{number}\mathrm{changement}\mathrm{either}\mathrm{error}\mathrm{function}$$$$\mathrm{for}\mathrm{the}\mathrm{further}\mathrm{integral}....$$$$$$

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