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

1 Answer
mason m
May 12, 2016

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

Explanation:

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#

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