Question

What is the antiderivative of e^x^2?

Answer 1

Answer: The antiderivative of $e^{x^{2}}$ is $\sqrt{\pi} \sqrt{e^{x^2}-1} +C$.

Explanation:

$I=\int{e^{x^2}dx}\\I^2=\int{e^{x^2}dx}\int{e^{x^2}dx}$

Using dawson integral.

$I^2=\int{e^{x^2}dx}\int{e^{y^2}dy}$

$I^2=\int{\int{e^{x^2+y^2}dxdy}}$

Put $x^2+y^2=r^2$

$I^2=\int{\int{e^{r^2}rdrd\theta}}$

$\int_{0}^{2\pi}{d\theta}\int_{0}^{x}{re^{r^2}dr}$

Use substitution method and sunbstitute $r^2=t$.

$\int_{0}^{2\pi}{d\theta}\int_{0}^{x^2}{\frac{1}{2}e^tdt}$

$I^2=\frac{1}{2}|\theta|_{0}^{2\pi}|e^t|_{0}^{x^2}\\I^2=\frac{1}{2}(2\pi)(e^{x^2}-1)\\I=\sqrt{\pi}\sqrt{e^{x^2}-1}$.

Therefore, the antiderivative of $e^{x^{2}}$ is $\sqrt{\pi} \sqrt{e^{x^2}-1} +C$.