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

1 Answer
Mar 19, 2018

Answer:

#-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) #