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Question Number 106391 by Ar Brandon last updated on 05/Aug/20

$$\int {\mathrm{e}}^{{2\mathrm{x}}^2+\mathrm{x}}\mathrm{dx}$$

Answered by Sarah85 last updated on 05/Aug/20

$$t=\sqrt{2}(x+\frac{1}{4})\iff x=\frac{\sqrt{2}}{2}t-\frac{1}{4}\Rightarrow dx=\frac{\sqrt{2}}{2}dt$$$$\frac{\sqrt{2}}{2}\int {\mathrm{e}}^{t^2-\frac{1}{8}}dt=\frac{\sqrt{2}}{{2\mathrm{e}}^{1/8}}\int {\mathrm{e}}^{t^2}dt=\frac{\sqrt{2\pi }}{{4\mathrm{e}}^{1/8}}\int \frac{{2\mathrm{e}}^{t^2}}{\sqrt{\pi }}dt=$$$$=\frac{\sqrt{2\pi }}{{4\mathrm{e}}^{1/8}}\mathrm{erfi}t=\frac{\sqrt{2\pi }}{{4\mathrm{e}}^{1/8}}\mathrm{erfi}\left(\sqrt{2}(x+\frac{1}{4})\right)+C$$

Commented by Ar Brandon last updated on 05/Aug/20

Thanks

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