If $\dfrac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}} = \dfrac{a}{b}$, then x equals:

1. $\dfrac{a^2}{a^2 + b^2}$

2. $\dfrac{b^2}{a+b}$

3. $\dfrac{ab}{a^2 + b^2}$

4. $\dfrac{2ab}{a^2 + b^2}$

Given,

$\Rightarrow \dfrac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}} = \dfrac{a}{b} \\[1em] \Rightarrow b\Big(\sqrt{1+x} + \sqrt{1-x}\Big) = a\Big(\sqrt{1+x} - \sqrt{1-x}\Big) \\[1em] \Rightarrow b\sqrt{1+x} + b\sqrt{1-x} = a\sqrt{1+x} - a\sqrt{1-x} \\[1em] \Rightarrow b\sqrt{1+x} - a\sqrt{1+x} = -a\sqrt{1-x} - b\sqrt{1-x} \\[1em] \Rightarrow (b-a)\sqrt{1+x} = -(a+b)\sqrt{1-x} \\[1em] \text{Squaring on Both Sides,} \\[1em] \Rightarrow (b-a)^2(1+x) = (a+b)^2(1-x) \\[1em] \Rightarrow (b^2 - 2ab + a^2)(1+x) = (a^2 + 2ab + b^2)(1-x) \\[1em] \Rightarrow (a^2 + b^2 - 2ab)(1+x) = (a^2 + b^2 + 2ab)(1-x) \\[1em] \Rightarrow (a^2 + b^2 - 2ab) + (a^2 + b^2 - 2ab)x = (a^2 + b^2 + 2ab) - (a^2 + b^2 + 2ab)x \\[1em] \Rightarrow (a^2 + b^2 - 2ab) - (a^2 + b^2 + 2ab) = -(a^2 + b^2 - 2ab)x - (a^2 + b^2 + 2ab)x \\[1em] \Rightarrow a^2 + b^2 - 2ab - a^2 - b^2 - 2ab = - a^2x - b^2x + 2abx - a^2x - b^2x - 2abx \\[1em] \Rightarrow a^2 - a^2 - b^2 + b^2 - 2ab - 2ab = - a^2x - a^2x - b^2x - b^2x + 2abx - 2abx \\[1em] \Rightarrow -4ab = - 2a^2x - 2b^2x \\[1em] \Rightarrow -4ab = -2(a^2 + b^2)x \\[1em] \Rightarrow x = \dfrac{-4ab}{-2(a^2 + b^2)} \\[1em] \Rightarrow x = \dfrac{2ab}{a^2 + b^2}.$

Hence, option 4 is the correct option.