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,

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

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

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

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