# Question #f092d

May 2, 2017

$\frac{1}{2} \ln {x}^{2} + c$

#### Explanation:

We can use substitution to cancel out the $x$
$u = \ln x$
$\mathrm{du} = \frac{1}{x} \mathrm{dx}$
$\mathrm{dx} = x \mathrm{du}$

$\int \frac{u}{x} \setminus \cdot x \mathrm{du}$
$= \int u \mathrm{du}$
$= \frac{1}{2} {u}^{2} + c$
$= \frac{1}{2} \ln {x}^{2} + c$