Answer
$\displaystyle \frac{1}{6}(3x-2)\cdot|3x-2|+C$
Work Step by Step
$\displaystyle \int|3x-2|dx=$
Shortcut formula:
$\displaystyle \qquad\int|ax+b|dx =\displaystyle \frac{1}{2a}(ax+b)|ax+b|+C$
Apply with $ \left[\begin{array}{l}
a=3\\
b=-2
\end{array}\right]$
$=\displaystyle \frac{1}{2(3)}(3x-2)|3x-2|+C$
= $\displaystyle \frac{1}{6}(3x-2)\cdot|3x-2|+C$
