Answer
$\dfrac{x}{1-x}$
where $x\ne-1,0,1$
Work Step by Step
We simplify as follows:
$\dfrac{1/x+1}{1/x^2-1}=\dfrac{1/x+1}{(\dfrac{1}{x}-1)(\dfrac{1}{x}+1)} \\=\dfrac{1}{\dfrac{1}{x}-1}\\=\dfrac{x}{1-x}$
Note that $x\ne-1,0,1$ because the original function becomes undefined at these points (division by zero).
