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

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

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