How do you integrate $int lnx/x^2$ by parts?

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Answer 1 · 2017-07-30T19:43:09

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

$intuv'dx=uv-intu'vdx$

Here,

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

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

Therefore,

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

How do you integrate int lnx/x^2int lnx/x^2 by parts?