Answer
$\dfrac{x+1}{1-x}$
Work Step by Step
The given expression, $
\dfrac{\dfrac{1}{x-1}+1}{\dfrac{1}{x+1}-1}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{\dfrac{1+x-1}{x-1}}{\dfrac{1-(x+1)}{x+1}}
\\\\=
\dfrac{\dfrac{x}{x-1}}{\dfrac{1-x-1)}{x+1}}
\\\\=
\dfrac{\dfrac{x}{x-1}}{\dfrac{-x}{x+1}}
\\\\=
\dfrac{x}{x-1}\div\dfrac{-x}{x+1}
\\\\=
\dfrac{x}{x-1}\cdot\dfrac{x+1}{-x}
\\\\=
\dfrac{\cancel{x}}{x-1}\cdot\dfrac{x+1}{-\cancel{x}}
\\\\=
\dfrac{x+1}{-(x-1)}
\\\\=
\dfrac{x+1}{1-x}
.\end{array}
