Answer
$\frac{x}{(1-x)}$
Work Step by Step
$\dfrac{\frac{1}{x}+1}{\frac{1}{x^2}-1}=\dfrac{\frac{x+1}{x}}{\frac{1-x^2}{x^2}}=\frac{x+1}{x}\frac{x^2}{1-x^2}=(x+1)\frac{x}{(1+x)(1-x)}=\frac{x}{(1-x)}$
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