How do you integrate $ln(x)/x^2$ ?

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Answer 1 · 2015-06-29T05:51:48

This can be rewritten as:

$int lnx * 1/x^2dx$

Using integration by parts, let:
$u = lnx$
$du = 1/xdx$
$dv = 1/x^2dx$
$v = -1/xdx$

With the formula
$uv - intvdu$

$=> -lnx/x - int -1/x*1/xdx$

$= -lnx/x - int -1/x^2dx$

$= color(blue)(-lnx/x - 1/x + C)$

How do you integrate ln(x)/x^2ln(x)/x^2?