The value of the integral ∫ - 2 2 x 3 + x e x x + 1 d x is equal to

Given,

$f\left(x\right)=\frac{\left|x^3+x\right|}{\left(e^{x\left|x\right|}+1\right)}dx$

and ${\int }_{-2}^{2}f\left(x\right)dx={\int }_0^2\left(f\left(x\right)+f\left(-x\right)\right)dx$

$={\int }_0^2\left(\frac{\left|x^3+x\right|}{\left(e^{x\left|x\right|}+1\right)}+\frac{\left|-x^3-x\right|}{\left(e^{-x\left|-x\right|}+1\right)}\right)dx$

$={\int }_0^2\left(\frac{\left|x^3+x\right|}{\left(e^{x\left|x\right|}+1\right)}+\frac{\left|x^3+x\right|}{\left(e^{-x\left|x\right|}+1\right)}\right)dx$

$={\int }_0^2\left(\frac{x^3+x}{\left(e^{x^2}+1\right)}+\frac{x^3+x}{\left(e^{-x^2}+1\right)}\right)dx$

$I={\int }_0^2\left(\frac{x^3+x}{1+e^{x^2}}+\frac{e^{x^2}\left(x^3+x\right)}{1+e^{x^2}}\right)dx$

$={\int }_0^2\left(x^3+x\right)dx$

$={\left[\frac{x^4}{4}+\frac{x^2}{2}\right]}_0^2$

$=4+2=6$