How do you evaluate the integral of $int x^3 lnx dx$ ?

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Answer 1 · 2016-05-12T01:07:40

$intx^3lnxdx=(x^4(4lnx-1))/16+C$

Use integration by parts, which states that:

$intudv=uv-intvdu$

So, for $intx^3lnxdx$ , let $u=lnx$ and $dv=x^3dx$ .

These imply that $du=1/xdx$ and $v=x^4/4$ (obtain these by differentiating $u$ and integrating $dv$ , respectively).

Plugging these into the integration by parts formula, this yields:

$intx^3lnxdx=lnx(x^4/4)-int(x^4/4)(1/x)dx$

$=(x^4lnx)/4-1/4intx^3dx$

$=(x^4lnx)/4-x^4/16+C$

$=(x^4(4lnx-1))/16+C$

How do you evaluate the integral of int x^3 lnx dxint x^3 lnx dx?