#### Answer

$\dfrac{3}{x-1}$

#### Work Step by Step

$\dfrac{6}{x^2-1}+\dfrac{3}{x+1} = \dfrac{6}{(x+1)(x-1)}+\dfrac{3}{x+1} = \dfrac{6(x+1)+3(x+1)(x-1)}{(x+1)(x-1)(x+1)} = \dfrac{(x+1)[6+3(x-1)]}{(x+1)(x-1)(x+1)} = \dfrac{6+3(x-1)}{(x+1)(x-1)} = \dfrac{6+3x-3)}{(x+1)(x-1)} = \dfrac{3x+3}{(x+1)(x-1)} = \dfrac{3(x+1)}{(x+1)(x-1)} = \dfrac{3}{x-1}$