# What is the integral of e^(-x^2)?

Mar 19, 2018

$- \frac{1}{2 x} \int \left(- 2 x \cdot {e}^{- {x}^{2}}\right) = - \frac{1}{2 x} \cdot \left({e}^{- {x}^{2}} + \lambda\right) , \lambda \in \mathbb{R}$

#### Explanation:

First We should change the expression into our int.
We know that $\int \left(u ' {e}^{u}\right) = {e}^{u} + \lambda , \lambda \in \mathbb{R} .$
$\int \left({e}^{- {x}^{2}}\right) = - \frac{1}{2 x} \int \left(- 2 x \cdot {e}^{- {x}^{2}}\right)$
Now We can integrate it properly. (here, $u = - {x}^{2} \iff u ' = - 2 x$)

$- \frac{1}{2 x} \int \left(- 2 x \cdot {e}^{- {x}^{2}}\right) = - \frac{1}{2 x} \cdot \left({e}^{- {x}^{2}} + \lambda\right)$