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

Given : $$x-\frac{1}{x}=1$$

=> $$\frac{x^2-1}{x}=1$$

=> $$x^2-1=x$$ ------------(i)

=> $$x^2-x-1=0$$

=> $$x=\frac{1\pm\sqrt5}{2}$$ -----------(ii)

To find : $$\frac{1}{x}(\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x^2+1}-\frac{1}{x^2-1})$$

= $$\frac{1}{x}[\frac{(x+1)-(x-1)}{(x+1)(x-1)}+\frac{(x^2-1)-(x^2+1)}{(x^2-1)(x^2+1)}]$$

= $$\frac{1}{x}[\frac{2}{x^2-1}+\frac{-2}{(x^2-1)(x^2+1)}]$$

= $$\frac{1}{x}[\frac{2}{x}-\frac{2}{x(x^2+1)}]$$      [Using equation (i)]

= $$\frac{2}{x^2}(1-\frac{1}{x^2+1})$$

= $$\frac{2}{x^2}(\frac{x^2+1-1}{x^2+1})$$

= $$\frac{2}{x^2}\times\frac{x^2}{x^2+1}=\frac{2}{x^2+1}$$

Adding '2' on both sides in equation (i), we get $$x^2+1=x+2$$

= $$\frac{2}{x+2}$$

Now substituting value from equation (ii) in above equation,

= $$\frac{4}{5\pm\sqrt5}$$

=> Ans - (C)