The value of the integral ∫ _ - 2 ^ 2 | x ^ 3 + x| ( e ^ x|x| + 1)...

Given,

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

and ${∫}_{- 2}^{2}f\left(x\right)dx = {∫}_{0}^{2}\left(f\left(x\right) + f\left(- x\right)\right)dx$

$= {∫}_{0}^{2}\left(\dfrac{\left|{x}^{3} + x\right|}{\left({e}^{x\left|x\right|} + 1\right)} + \dfrac{\left|- {x}^{3} - x\right|}{\left({e}^{- x\left|- x\right|} + 1\right)}\right)dx$

$= {∫}_{0}^{2}\left(\dfrac{\left|{x}^{3} + x\right|}{\left({e}^{x\left|x\right|} + 1\right)} + \dfrac{\left|{x}^{3} + x\right|}{\left({e}^{- x\left|x\right|} + 1\right)}\right)dx$

$= {∫}_{0}^{2}\left(\dfrac{{x}^{3} + x}{\left({e}^{{x}^{2}} + 1\right)} + \dfrac{{x}^{3} + x}{\left({e}^{- {x}^{2}} + 1\right)}\right)dx$

$I = {∫}_{0}^{2}\left(\dfrac{{x}^{3} + x}{1 + {e}^{{x}^{2}}} + \dfrac{{e}^{{x}^{2}}\left({x}^{3} + x\right)}{1 + {e}^{{x}^{2}}}\right)dx$

$= {∫}_{0}^{2}\left({x}^{3} + x\right)dx$

$= {\left[\dfrac{{x}^{4}}{4} + \dfrac{{x}^{2}}{2}\right]}_{0}^{2}$

$= 4 + 2 = 6$