What is the value of $\frac{1+x^2}{1-x}÷\frac{1-x^4}{x-1}×\frac{x(1-x)}{(1+x)}$?

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

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

$\frac{1+x^2}{1-x}÷\frac{1-x^4}{x-1}×\frac{x(1-x)}{(1+x)}$ \(\frac{1+x^2}{1-x}\) × \(\frac{-1(1-x)}{(1+x^2)(1- x^2)}\) × \(\frac{x(1-x)}{(1+x)}\) = $\frac{-x}{(1+x)^2}$