Answer: The antiderivative of [tex]e^{x^{2}}[/tex] is [tex]\sqrt{\pi} \sqrt{e^{x^2}-1} +C[/tex].
Explanation:
[tex]I=\int{e^{x^2}dx}\\I^2=\int{e^{x^2}dx}\int{e^{x^2}dx}[/tex]
Using dawson integral.
[tex]I^2=\int{e^{x^2}dx}\int{e^{y^2}dy}[/tex]
[tex]I^2=\int{\int{e^{x^2+y^2}dxdy}}[/tex]
Put [tex]x^2+y^2=r^2[/tex]
[tex]I^2=\int{\int{e^{r^2}rdrd\theta}}[/tex]
[tex]\int_{0}^{2\pi}{d\theta}\int_{0}^{x}{re^{r^2}dr}[/tex]
Use substitution method and sunbstitute [tex]r^2=t[/tex].
[tex]\int_{0}^{2\pi}{d\theta}\int_{0}^{x^2}{\frac{1}{2}e^tdt}[/tex]
[tex]I^2=\frac{1}{2}|\theta|_{0}^{2\pi}|e^t|_{0}^{x^2}\\I^2=\frac{1}{2}(2\pi)(e^{x^2}-1)\\I=\sqrt{\pi}\sqrt{e^{x^2}-1}[/tex].
Therefore, the antiderivative of [tex]e^{x^{2}}[/tex] is [tex]\sqrt{\pi} \sqrt{e^{x^2}-1} +C[/tex].
