Answer
$\frac{4x+1}{(x+1)(x-1)}$
Work Step by Step
$\frac{2}{x+1}+\frac{2}{x-1}+\frac{1}{x^2-1}=\frac{2(x-1)}{(x+1)(x-1)}+\frac{2(x+1)}{(x+1)(x-1)}+\frac{1}{(x+1)(x-1)}=\frac{2(x-1)+2(x+1)+1}{(x+1)(x-1)}=\frac{2x-2+2x+2+1}{(x+1)(x-1)}=\frac{4x+1}{(x+1)(x-1)}$
