Answer
$-\frac{(x+1) }{(x-1) }$.
Work Step by Step
The given expression is
$=\frac{\frac{1}{x-1}+1}{\frac{1}{x+1}-1}$
Multiply the numerator and the denominator by $(x-1)(x+1)$.
$=\frac{(x-1)(x+1)}{(x-1)(x+1)}\cdot \frac{\frac{1}{x-1}+1}{\frac{1}{x+1}-1}$
Use the distributive property.
$=\frac{(x-1)(x+1) \cdot \frac{1}{x-1}+(x-1)(x+1) }{(x-1)(x+1) \cdot \frac{1}{x+1}-(x-1)(x+1) }$
Simplify.
$=\frac{(x+1) +(x-1)(x+1) }{(x-1)-(x-1)(x+1) }$
Factor out common terms in the numerator and denominator.
$=\frac{(x+1) [1+(x-1)] }{(x-1)[1-(x+1)] }$
$=\frac{(x+1) [1+x-1] }{(x-1)[1-x-1] }$
$=\frac{(x+1) [x] }{(x-1)[-x] }$
Cancel common terms.
$=\frac{(x+1) }{-(x-1) }$
$=-\frac{(x+1) }{(x-1) }$.
