Integral of $$$e^{x^{2}} - 1$$$
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Find $$$\int \left(e^{x^{2}} - 1\right)\, dx$$$.
Solution
Integrate term by term:
$${\color{red}{\int{\left(e^{x^{2}} - 1\right)d x}}} = {\color{red}{\left(- \int{1 d x} + \int{e^{x^{2}} d x}\right)}}$$
Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:
$$\int{e^{x^{2}} d x} - {\color{red}{\int{1 d x}}} = \int{e^{x^{2}} d x} - {\color{red}{x}}$$
This integral (Imaginary Error Function) does not have a closed form:
$$- x + {\color{red}{\int{e^{x^{2}} d x}}} = - x + {\color{red}{\left(\frac{\sqrt{\pi} \operatorname{erfi}{\left(x \right)}}{2}\right)}}$$
Therefore,
$$\int{\left(e^{x^{2}} - 1\right)d x} = - x + \frac{\sqrt{\pi} \operatorname{erfi}{\left(x \right)}}{2}$$
Add the constant of integration:
$$\int{\left(e^{x^{2}} - 1\right)d x} = - x + \frac{\sqrt{\pi} \operatorname{erfi}{\left(x \right)}}{2}+C$$
Answer
$$$\int \left(e^{x^{2}} - 1\right)\, dx = \left(- x + \frac{\sqrt{\pi} \operatorname{erfi}{\left(x \right)}}{2}\right) + C$$$A