Question #a8dd2

1 Answer
Nov 8, 2016

#intx^3e^(2x^2)dx=1/8e^(2x^2)(2x^2-1)+C#

Explanation:

#I=intx^3e^(2x^2)dx#

Before using integration by parts (IBP), we can make a simpler substitution. Let #t=2x^2# so that #dt=4xdx#. This also implies that #x^2=1/2t#.

Modifying the integral:

#I=1/4intx^2e^(2x^2)(4xdx)=1/4int1/2te^tdt=1/8intte^tdt#

Now we should apply IBP. This integration technique takes the form #intudv=uv-intvdu#. So, for #intte^tdt#, we should let:

#{(u=t,=>,du=dt),(dv=e^tdt,=>,v=e^t):}#

Recall to differentiate #u# and integrate #dv#.

Thus:

#I=1/8(te^t-inte^tdt)=1/8(te^t-e^t)=1/8e^t(t-1)#

Since #t=2x^2# this becomes:

#I=1/8e^(2x^2)(2x^2-1)+C#