How do you integrate $int (lnx)^2/x^3$ using integration by parts?

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Answer 1 · 2017-07-30T18:35:20

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

The integration by oarts is

$intuv'dx=uv-intu'vdx$

Here,

$u=(lnx)^2$ , $=>$ , $u'=2lnx*1/x$

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

Therefore,

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

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

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

So,

$int(lnx)/x^3=-lnx/(2x^2)+intdx/(2x^3)=-lnx/(2x^2)-1/(4x^2)$

Therefore,

$int((lnx)^2dx)/x^3=-(lnx)^2/(2x^2)-lnx/(2x^2)-1/(4x^2)+C$

How do you integrate int (lnx)^2/x^3int (lnx)^2/x^3 using integration by parts?