How can you integrate #xe^(x^2) dx# using integration by parts?

1 Answer
May 19, 2015

We do not need to integrate by parts, but since that is what is specified:

We can integrate #xe^(x^2)dx# by sustitution #w=x^2# and we end up with

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

Therefore, to use parts, I will choose #u=1# and #dv=xe^(x^2)dx#

This makes #du=0dx# and #v= 1/2 e^(x^2)#

The parts formula gives us;

#1/2e^(x^2)- int 0 dx#

#= 1/2 e^(x^2) + C#