What is the antiderivative of e^x^2?

Question

asked Oct 17, 2019 73.1k views

1 Answer

Cieunteung · Answer 1 · 2019-10-22T14:25:21+0000

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

Step-by-step 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 .

$\int_(0)^(2\pi){d\theta}\int_(0)^(x^2){(1)/(2)e^tdt}$

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

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