Answer
The equation has no solution.
Work Step by Step
$\dfrac{x}{x-1}-\dfrac{1}{x+1}=\dfrac{2}{x^{2}-1}$
Factor the denominator of the fraction on the right side of the equation:
$\dfrac{x}{x-1}-\dfrac{1}{x+1}=\dfrac{2}{(x-1)(x+1)}$
Multiply the whole equation by $(x-1)(x+1)$:
$(x-1)(x+1)\Big[\dfrac{x}{x-1}-\dfrac{1}{x+1}=\dfrac{2}{(x-1)(x+1)}\Big]$
$x(x+1)-(x-1)=2$
Evaluate the indicated operations:
$x^{2}+x-x+1=2$
Take $2$ to the left side and simplify:
$x^{2}+x-x+1-2=0$
$x^{2}-1=0$
Solve by factoring:
$(x-1)(x+1)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x-1=0$
$x=1$
$x+1=0$
$x=-1$
The original equation is undefined for $x=1$ and $x=-1$, so it has no solution.
