$f(x)=\frac{\left|x^{3}+x\right|}{\left(e^{x|x|}+1\right)} d x$
$\int\limits_{-2}^{2} f(x) d x=\int_{0}^{2}(f(x)+f(-x)) d x$
$=\int\limits_{0}^{2}\left(\frac{\left|x^{3}+x\right|}{\left(e^{x|x|}+1\right)}+\frac{\left|-x^{3}-x\right|}{\left(e^{-x|-x|}+1\right)}\right) d x$
$=\int\limits_{0}^{2}\left(\frac{\left|x^{3}+x\right|}{\left(e^{x|x|}+1\right)}+\frac{\left|x^{3}+x\right|}{\left(e^{-x|x|}+1\right)}\right) d x$
$=\int\limits_{0}^{2}\left(\frac{x^{3}+x}{\left(e^{x^{2}}+1\right)}+\frac{x^{3}+x}{\left(e^{-x^{2}}+1\right)}\right) d x$
$I=\int\limits_{0}^{2}\left(\frac{x^{3}+x}{1+e^{x^{2}}}+\frac{e^{x^{2}}\left(x^{3}+x\right)}{1+e^{x^{2}}}\right) d x$
$=\int\limits_{0}^{2}\left(x^{3}+x\right) d x$
$=\left[\frac{x^{4}}{4}+\frac{x^{2}}{2}\right]_{0}^{2}$
$=4+2=6$