Answer
$-3$
Work Step by Step
$\frac{x}{x-1} + \frac{1}{x+1} = \frac{2}{x^{2}-1}$
Using $(a^{2}-b^{2}) = (a+b)(a-b)$ formula,
$x^{2}-1 = (x^{2}-1^{2}) = (x+1)(x-1)$
$\frac{x}{x-1} + \frac{1}{x+1} = \frac{2}{(x+1)(x-1)}$
$x=1 $ or $x=-1$ makes the denominator zero. So,
$\frac{x}{x-1} + \frac{1}{x+1} = \frac{2}{(x+1)(x-1)} ; $$ x\ne -1, $ or $x \ne 1$
Multiply both sides by $(x+1)(x-1)$ to clear fractions.
$(x+1)(x-1)(\frac{x}{x-1} + \frac{1}{x+1} )=(x+1)(x-1)( \frac{2}{(x+1)(x-1)}) ; $$ x\ne -1, $ or $x \ne 1$
$(x+1)x+(x-1) = 2 ; x\ne -1, $ or $x \ne 1$
$x^{2}+x+x-1-2=0 $
$x^{2}+2x-3=0 $
By factoring,
$(x+3)(x-1)=0$
$x=-3$ or $x=1$
$x = 1$ makes the denominator zero, so
$x=-3$ is the solution.