The value of the integral $$\int_{-2}^{2} \frac{|x^3+x|}{(e^{x|x|}+1)} dx$$ is equal to

We need to evaluate $$\int_{-2}^{2} \frac{|x^3+x|}{e^{x|x|}+1} dx$$. First, observe that $$x^3 + x = x(x^2 + 1)$$ and since $$x^2 + 1 > 0$$ always, the sign of $$x^3 + x$$ depends solely on $$x$$. For $$x \geq 0$$ we have $$|x^3 + x| = x^3 + x$$, while for $$x < 0$$ it follows that $$|x^3 + x| = -(x^3 + x) = -x^3 - x$$. Similarly, note that $$x|x| = x^2$$ when $$x \geq 0$$ and $$x|x| = -x^2$$ when $$x < 0$$.

From these observations, we split the integral into two parts:

$$I = \int_{-2}^{2} \frac{|x^3+x|}{e^{x|x|}+1} dx = \int_{-2}^{0} \frac{-(x^3+x)}{e^{-x^2}+1} dx + \int_{0}^{2} \frac{x^3+x}{e^{x^2}+1} dx.$$ Substituting $$x \to -x$$ in the first integral gives $$\int_{-2}^{0} \frac{-(x^3+x)}{e^{-x^2}+1} dx = \int_{0}^{2} \frac{x^3+x}{e^{-x^2}+1} dx,$$ so that $$I = \int_{0}^{2} (x^3+x)\left[\frac{1}{e^{-x^2}+1} + \frac{1}{e^{x^2}+1}\right] dx.$$

Since $$\frac{1}{e^{-x^2}+1} + \frac{1}{e^{x^2}+1} = \frac{e^{x^2}}{e^{x^2}+1} + \frac{1}{e^{x^2}+1} = \frac{e^{x^2}+1}{e^{x^2}+1} = 1,$$ the integral reduces to $$I = \int_{0}^{2} (x^3+x)\, dx = \left[\frac{x^4}{4} + \frac{x^2}{2}\right]_0^2 = \frac{16}{4} + \frac{4}{2} = 4 + 2 = 6.$$

The answer is Option D: $$6$$.