#### Answer

$x=-3$

#### Work Step by Step

$\dfrac{2}{x+1}-\dfrac{1}{x-1}=\dfrac{2x}{x^{2}-1}$
Factor the fraction on the right side completely:
$\dfrac{2}{x+1}-\dfrac{1}{x-1}=\dfrac{2x}{(x-1)(x+1)}$
Multipy the whole equation by $(x-1)(x+1)$:
$(x-1)(x+1)\Big[\dfrac{2}{x+1}-\dfrac{1}{x-1}=\dfrac{2x}{(x-1)(x+1)}\Big]$
$2(x-1)-(x+1)=2x$
$2x-2-x-1=2x$
Take all terms with $x$ to the left side and all terms without $x$ to the right side. Then, solve for $x$:
$2x-2x-x=1+2$
$-x=3$
$x=-3$