Answer
$\displaystyle \frac{2}{x (x+1)} $,$\qquad x\neq 0,\ \pm 1,\ -\displaystyle \frac{1}{3}$
Work Step by Step
Division = Multiplication with the reciprocal of the divisor,
$... =\displaystyle \frac{6x+2}{x^{2}-1}\cdot\frac{x-1}{3x^{2}+x}$
Before multiplying, we factor what we can, so we can cancel common factors.
$6x+2=2(3x+1),$
$ x^{2}-1$= difference of squares = $(x+1)(x-1)$
$3x^{2}+x=x(3x+1)$
Problem $= \displaystyle \frac{2(3x+1)\cdot(x-1)}{ (x+1)(x-1)\cdot x(3x+1)},\qquad x\neq 0, \pm 1,\ -\displaystyle \frac{1}{3}$
All numbers producing zero in the denominator are excluded from the domain.
There are common factors; reduce the expression to:
= $\displaystyle \frac{2}{x (x+1)} $,$\qquad x\neq 0,\ \pm 1,\ -\displaystyle \frac{1}{3}$
