How do solve $int lnx/x^3 dx$ ?

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Answer 1 · 2016-12-24T07:41:29

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

We need

$intx^ndx=x^(n+1)/(n+1)+C(n!=-1)$

$(lnx)'=1/x$

$intuv'dx=uv-intu'vdx$

Here,

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

$v'=1/x^3$ , $=>$ , $v=-1/(2x^2)$

Therefore,

$int(lnxdx)/x^3=-lnx/(2x^2)+1/2intdx/x^3$

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

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

How do solve int lnx/x^3 dxint lnx/x^3 dx ?