How do you integrate #int x(lnx)^2 # using integration by parts?

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Answer 1 · 2018-05-20T14:27:30

The answer is #=x^2/2(lnx)^2-x^2/2lnx+x^2/4+C#

#intuv'=uv-intu'v#

Here,

#u=(lnx)^2#, #=>#, #du=2/xlnx#

#v'=x#, #=>#, #v=x^2/2#

Therefore, the integral is

#intx(lnx)^2dx=x^2/2(lnx)^2-int2/xlnx*x^2/2dx#

#=x^2/2(lnx)^2-intxlnxdx#

To calculate #intxlnxdx#, perform the integration by parts

#u=lnx#, #=>#, #u'=1/x#

#v'=x#, #=>#, #v=x^2/2#

Therefore,

#intxlnxdx=x^2/2lnx-int1/x*x^2/2dx#

#=x^2/2lnx-1/2intxdx#

#=x^2/2lnx-x^2/4#

Putting all together

#intx(lnx)^2dx=x^2/2(lnx)^2-x^2/2lnx+x^2/4+C#