#### Answer

$x=-5$

#### Work Step by Step

$\frac{x}{x-2}+\frac{1}{x+2}=\frac{8}{x^{2}-4}$
We multiply through by $(x-2)(x+2)$:
$x(x+2)+(x-2)=8$
Next we distribute:
$x^{2}+2x+x-2=8$
$x^{2}+3x-10=0$
And factor:
$(x-2)(x+5)=0 $
$x-2=0$ or $x+5=0$
$x=2$ or $x=-5$
However, we have to eliminate $x=2$ because it creates division by 0 in the original equation. So the only solution is $x=-5$.