Question

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

Answer 1

`-1/(2x)int(-2x*e^(-x^2))=-1/(2x)*(e^(-x^2)+lambda), lambda in RR`

Explanation:

First We should change the expression into our int.
We know that `int(u'e^u) =e^u+lambda, lambda in RR.`
`int(e^(-x^2)) =-1/(2x)int(-2x*e^(-x^2))`
Now We can integrate it properly. (here, `u=-x^2 <=> u'=-2x`)

`-1/(2x)int(-2x*e^(-x^2))=-1/(2x)*(e^(-x^2)+lambda)`