If $$x - \frac{1}{x} = 1,$$   then what is the value of $$\frac{1}{x}\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}} + \frac{1}{{{x^2} + 1}} - \frac{1}{{{x^2} - 1}}} \right)?$$

$$\eqalign{ & x - \frac{1}{x} = 1 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 1 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} - 2 = 3 \cr & \Rightarrow x + \frac{1}{x} = \pm \sqrt 5 \cr & \frac{1}{x}\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}} + \frac{1}{{{x^2} + 1}} - \frac{1}{{{x^2} - 1}}} \right) \cr & \Rightarrow \frac{1}{x}\left( {\frac{2}{{{x^2} - 1}} - \frac{2}{{\left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right)}}} \right) \cr & \Rightarrow \frac{1}{x}\left( {\frac{{2{x^2} + 2 - 2}}{{\left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right)}}} \right) \cr & \because \,{x^2} - 1 = x,\,{x^2} + 1 = \sqrt {5x} \cr & \Rightarrow \frac{1}{x}\left( {\frac{{2{x^2}}}{{x \times \sqrt {5x} }}} \right) \cr & \Rightarrow \pm \frac{2}{{\sqrt {5x} }} \cr} $$