Indefinite Integral of the Form e^{x^2}
Theorem
$$ \int e^{x^2}dx = \sum\limits_{n=0}^\infty \dfrac{x^{2n+1}}{(2n+1)n!}+C $$
Explanation
Just like the form $e^{-x^{2}}$, it is difficult to integrate using general methods. There is a method to integrate by defining the error function, imaginary error function, erfi, but this article introduces solving it using Taylor series expansion.
Proof
By the method of Taylor series expansion,
$$ e^{x} = \sum\limits_{n=0}^{\infty} \dfrac{x^{n}}{n!} = 1 + x + \dfrac{x^{2}}{2!} + \cdots + \dfrac{x^{n}}{n!} + \cdots $$
Inserting $x^2$ instead of $x$ gives
$$ e^{x^2}=\sum\limits_{n=0}^\infty \dfrac{x^{2n}}{n!}=1+x^2+\dfrac{x^4}{2!}+\cdots +\dfrac{x^{2n}}{n!}+\cdots $$
Integrating both sides indefinitely gives
$$ \begin{align*} \int e^{x^2}dx =&\ \int (1+x^2+\dfrac{x^4}{2!}+\cdots +\dfrac{x^{2n}}{n!}+\cdots)dx \\ =&\ x + \dfrac{x^3}{3 \cdot 1!} + \dfrac{x^5}{5 \cdot 2!}+\dfrac{x^7}{7 \cdot 3!}+ \cdots +\dfrac{x^{2n+1}}{(2n+1)n!}+\cdots +C \end{align*} $$
Therefore
$$ \int e^{x^2}dx=\sum \limits_{n=0}^\infty \dfrac{x^{2n+1}}{(2n+1)n!}+C $$
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