# What is the antiderivative of  lnx/x^2?

Sep 11, 2016

If we take this to be integration by parts, then, let:

$u = \ln x$
$\mathrm{du} = \frac{1}{x} \mathrm{dx}$
$\mathrm{dv} = \frac{1}{x} ^ 2 \mathrm{dx}$
$v = - \frac{1}{x}$

$u v - \int v \mathrm{du}$

$= - \frac{\ln x}{x} - \left(- \int \frac{1}{x} ^ 2 \mathrm{dx}\right)$

$= \textcolor{b l u e}{- \frac{\ln x}{x} - \frac{1}{x} + C}$