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Answer 1 · 2016-11-08T14:23:02

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

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

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:

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