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

Jun 29, 2015

This can be rewritten as:

$\int \ln x \cdot \frac{1}{x} ^ 2 \mathrm{dx}$

Using integration by parts, let:
$u = \ln x$
$\mathrm{du} = \frac{1}{x} \mathrm{dx}$
$\mathrm{dv} = \frac{1}{x} ^ 2 \mathrm{dx}$
$v = - \frac{1}{x} \mathrm{dx}$

With the formula
$u v - \int v \mathrm{du}$

$\implies - \ln \frac{x}{x} - \int - \frac{1}{x} \cdot \frac{1}{x} \mathrm{dx}$

$= - \ln \frac{x}{x} - \int - \frac{1}{x} ^ 2 \mathrm{dx}$

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