How do you integrate #int x^3e^(x^2)# by integration by parts method?

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Answer 1 · 2016-10-25T18:00:56

The integral is #(x^2-1)/2e^(x^2)+C#

First we use the substitution #u=x^2#
so #du=2xdx#
So the integral becomes #intx^3e^(x^2)dx#
#=1/2intue^udu#

This is the integration by parts
let #p=u# then #p'=1#
and #v'=e^u# then #v=e^u#

#intpv'=pv-intp'v#

#1/2intue^udu=1/2(ue^u-inte^udu)#
#=1/2(ue^u-e^u)#

#=(u-1)/2e^u#

Going back to x

#intx^3e^(x^2)dx=(x^2-1)/2e^(x^2) +C#