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#