How do you integrate $(x^2 ln x) / x$ ?

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Answer 1 · 2016-06-08T03:55:59

$(x^2(2lnx-1))/4+C$

First note that one pair of $x$ terms will cancel:

$(x^2lnx)/x=xlnx$

So, we want to find:

$intxlnxdx$

We will use integration by parts, which takes the form:

$intudv=uv-intvdu$

So, for $intxlnxdx$ , let:

${(u=lnx),(dv=xdx):}$

Differentiate $u$ and integrate $dv$ to show that:

${(du=1/xdx),(v=x^2/2):}$

Plugging these back into the integration by parts formula:

$intxlnxdx=(x^2lnx)/2-intx^2/2(1/x)dx$

$=(x^2lnx)/2-1/2intxdx$

$=(x^2lnx)/2-x^2/4+C$

$=(x^2(2lnx-1))/4+C$

How do you integrate (x^2 ln x) / x(x^2 ln x) / x?