Given, $f(x)=\max \{x+|x|, x-[x]\}$
$=\begin{cases}2 x, & x \geq 0 \\ x-[x], & x \leq 0\end{cases}$
$\therefore \int\limits_{-3}^{3} f(x) d x=\int\limits_{-3}^{0} x-[x] d x+\int\limits_{0}^{3} 2 x\, d x$
$=3 \int\limits_{-1}^{0}(1+x) d x+2 \int\limits_{0}^{3} x\, d x$
$[\because x-[x]$ is a periodic function at $x=1]$
$=3\left[x+\frac{x^{2}}{2}\right]_{-1}^{0}+2\left[\frac{x^{2}}{2}\right]_{0}^{3}$
$=3\left[0-0-\left(-1+\frac{1}{2}\right)\right]+3^{2}-0$
$=3\left[\frac{1}{2}\right]+9=\frac{3}{2}+9$
$=\frac{21}{2}$