What is the integral of #int (x^2)*e^(x^2) dx #?

Question

17557 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-24T19:30:31

# 1/2xe^(x^2)-sqrt(pi)/4"erfi"(x)#

#d/dx(xe^(x^2))= 2x^2e^(x^2)+e^(x^2)#

then

#int x^2 e^(x^2)dx = 1/2(xe^(x^2)-int e^(x^2)dx)#

but #"erfi"(x) = 2/sqrt(pi)int_0^x e^(x^2)dx# so

#int x^2 e^(x^2)dx = 1/2xe^(x^2)-sqrt(pi)/4"erfi"(x)#

NOTE:

#"erfi"(x)# or error function is defined as

#"erfi"(x) = 2/sqrtpi int_0^x e^(-xi^2) d xi#