How do you integrate #int 3 xln x^2 dx # using integration by parts?

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Answer 1 · 2017-02-12T10:23:23

The answer is #=3x^2ln(|x|)-3/2x^2+C#

#intu'vdx=uv-intuv'dx#

Here, we have

#int3xln(x^2)dx=6intxlnxdx#

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

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

Therefore,

#int3xln(x^2)dx=6(1/2x^2lnx-int1/2xdx)#

#=3x^2lnx-3*1/2*x^2#

#=3x^2ln(|x|)-3/2x^2+C#