(1+(1/(x^2-1)))/((1/x)-(x/(x+1)))
Thank you for using brackets, but this is still very hard to read...
Is it $\displaystyle \frac{1 + \frac{1}{x^2 - 1}}{\frac{1}{x} - \frac{x}{x + 1}}$?
If so, start by making some common denominators
$\displaystyle \frac{1 + \frac{1}{x^2 - 1}}{\frac{1}{x} - \frac{x}{x + 1}} = \frac{\frac{x^2 - 1}{x^2 - 1} + \frac{1}{x^2 - 1}}{\frac{x + 1}{x(x + 1)} - \frac{x^2}{x(x + 1)}}$
$\displaystyle = \frac{\frac{x^2}{x^2 - 1}}{\frac{-x^2 + x + 1}{x(x + 1)}}$
$\displaystyle = \frac{\frac{x^2}{(x - 1)(x + 1)}}{\frac{-x^2 + x + 1}{x(x + 1)}}$
$\displaystyle = \frac{x^3(x + 1)}{(x - 1)(x + 1)(-x^2 + x + 1)}$
$\displaystyle = \frac{x^3}{(x - 1)(-x^2 + x + 1)}$.