The value of $$\displaystyle\int_{-1}^{1} \dfrac{\left(1 + \sqrt{|x| - x}\right)e^{-x} + \left(\sqrt{|x| - x}\right)e^{-x}}{e^{x} + e^{-x}} \, dx$$ is equal to

Consider the integral
$$I = \displaystyle\int_{-1}^{1}\dfrac{\bigl(1+\sqrt{|x|-x}\bigr)\,e^{-x}+\sqrt{|x|-x}\,e^{x}}{e^{-x}+e^{x}}\;dx$$

Step 1: Separate the numerator
Write the numerator as two separate parts:
$$\bigl(1+\sqrt{|x|-x}\bigr)e^{-x}+ \sqrt{|x|-x}\,e^{x} = e^{-x}+ \sqrt{|x|-x}\,(e^{-x}+e^{x}).$$

Step 2: Split the fraction
Substitute this expression back in the integrand and divide every term by the common denominator $$e^{-x}+e^{x}$$:
$$\dfrac{e^{-x}+ \sqrt{|x|-x}\,\bigl(e^{-x}+e^{x}\bigr)}{e^{-x}+e^{x}} =\dfrac{e^{-x}}{e^{-x}+e^{x}}+\sqrt{|x|-x}.$$

Step 3: Write the first term in a simpler form
Dividing both numerator and denominator of the first fraction by $$e^{-x}$$ gives
$$\dfrac{e^{-x}}{e^{-x}+e^{x}} =\dfrac{1}{1+e^{2x}}.$$

Thus the given integral becomes the sum of two separate integrals:
$$I=\underbrace{\int_{-1}^{1}\dfrac{dx}{1+e^{2x}}}_{I_1}\;+\; \underbrace{\int_{-1}^{1}\sqrt{|x|-x}\;dx}_{I_2}.$$

Step 4: Evaluate $$I_1$$
For every $$x\in[0,1]$$, use the substitution $$t=-x$$ in the interval $$[-1,0]$$:

$$\int_{-1}^{0}\dfrac{dx}{1+e^{2x}} =\int_{0}^{1}\dfrac{e^{2t}}{1+e^{2t}}\;dt.$$ Hence
$$I_1 =\int_{0}^{1}\left[\dfrac{1}{1+e^{2t}}+\dfrac{e^{2t}}{1+e^{2t}}\right]dt =\int_{0}^{1}1\,dt =1.$$

Step 5: Evaluate $$I_2$$
Note that $$|x|-x=0$$ for $$x\ge 0,$$ so the integrand is zero on $$[0,1].$$
For $$x\in[-1,0),$$ $$|x|=-x,$$ so $$|x|-x=-x-x=-2x\quad\Longrightarrow\quad \sqrt{|x|-x}=\sqrt{-2x}.$$

Therefore
$$I_2=\int_{-1}^{0}\sqrt{-2x}\;dx.$$ Put $$u=-x\;(u\in[0,1]),\;dx=-du:$$

$$I_2=\int_{1}^{0}\sqrt{2u}\;(-du) =\int_{0}^{1}\sqrt{2u}\;du =\sqrt{2}\int_{0}^{1}u^{1/2}\;du =\sqrt{2}\left[\dfrac{2}{3}u^{3/2}\right]_{0}^{1} =\dfrac{2\sqrt{2}}{3}.$$

Step 6: Combine the results
$$I=I_1+I_2 =1+\dfrac{2\sqrt{2}}{3}.$$

The value of the integral is $$1+\dfrac{2\sqrt{2}}{3}.$$
Therefore, the correct choice is Option D.