Answer
$=\displaystyle \frac{x(x+3)}{(x-2)(x-1)},\qquad x\neq -1, -2, -3,1,2$
Work Step by Step
Factor what we can:
$x^{2}+x=x(x+1)$
$x^{2}-4=(x-2)(x+2)\quad $(diff. of squares)
$x^{2}-1=(x-1)(x+1)\quad $(diff. of squares)
$ x^{2}+5x+6\quad$ ... find factors of c whose sum is b...
... we find 2 and 3 ...
$=(x+2)(x+3)$
Expression $=\displaystyle \frac{x(x+1)}{(x-2)(x+2)}\div\frac{(x-1)(x+1)}{(x+2)(x+3)}$
... Division with $\displaystyle \frac{a}{b}$ equals multiplication with $\displaystyle \frac{b}{a}.$
(neither a or b can be 0,
because of the denominator we exclude $-2$ and $-3$,
numerator: exclude $-1$ and $1$)
$=\displaystyle \frac{x(x+1)}{(x-2)(x+2)}\cdot\frac{(x+2)(x+3)}{(x-1)(x+1)}$
...exclude any additional values that yield 0 in the denominator:
$x\neq-1, -2, -3,1,2$
... cancel common factors
$=\displaystyle \frac{x(x+3)}{(x-2)(x-1)},\qquad x\neq -1, -2, -3,1,2$
