Answer
$\frac{x(x+3)}{(x-1)(x-2)}; x \ne 1,-1,2,-2,-3;$
Work Step by Step
$ \frac{(x^{2}+x)}{x^{2}-4} \div \frac{x^{2}-1}{x^{2}+5x+6}$
Factorize the numerator and the denominators.
$ =\frac{x(x+1)}{(x+2)(x-2)} \div \frac{(x+1)(x-1)}{(x+3)(x+2)}$
$x\ne 1,-1,2,-2,-3$ for non-zero denominators. So
$ =\frac{x(x+1)}{(x+2)(x-2)} \div \frac{(x+1)(x-1)}{(x+3)(x+2)};x\ne 1,-1,2,-2,-3;$
Invert the divisor and multiply.
$ =\frac{x(x+1)}{(x+2)(x-2)} \times \frac{(x+3)(x+2)}{(x+1)(x-1)};x\ne 1,-1,2,-2,-3;$
Divide out common factors.
$=\frac{x(x+3)}{(x-1)(x-2)}; x \ne 1,-1,2,-2,-3;$
