Answer
See below.
Work Step by Step
Our substitution will be based on the fact that $\displaystyle \frac{1}{2}x|x|$ is an antiderivative of $|x|$.
Check:
$\displaystyle \frac{d}{dx}[ \displaystyle \frac{1}{2}x\cdot|x|]=\frac{1}{2}\{\frac{d}{dx}[x]\cdot|x|+x\cdot\frac{d}{dx}[\ |x|\ ]\}\qquad$ product rule
$=\displaystyle \frac{1}{2}( |x|+x\cdot\frac{|x|}{x})\qquad $ see section 11.4, $\displaystyle \frac{d}{dx}[\ |x|\ ]$
$=\displaystyle \frac{1}{2}(2|x|)$
$=|x|$
so, $\displaystyle \int|x|dx= \displaystyle \frac{1}{2}x|x|+C.$
$\displaystyle \int|ax+b|dx=\qquad\quad \left[\begin{array}{ll}
u=ax+b, & du=adx, \\
& dx=\frac{1}{a}du
\end{array}\right]$
$=\displaystyle \frac{1}{a}\int|u|du$
$=\displaystyle \frac{1}{a}(\frac{1}{2}u|u|)+C$
bring back the variable $x$
$=\displaystyle \frac{1}{2a}(ax+b)|ax+b|+C$
