You should consider the fact that |x| is one linear function for \(\displaystyle{x}\in{\left(-\infty;{0}\right]}\) and a linear function for \(x\in[0;\infty)\) to break the interval of integration into subintervals that are easier.

\(\displaystyle{\int_{{-{2}}}^{{{3}}}}{\left|{x}\right|}{\left.{d}{x}\right.}={\int_{{-{2}}}^{{{0}}}}-{x}{\left.{d}{x}\right.}+{\int_{{{0}}}^{{{3}}}}{x}{\left.{d}{x}\right.}\)