Answer
$\frac{(x+4)(x+2)}{(x-5)}; x \ne -6,-3,-1,3,5$
Work Step by Step
Given expression is
$\frac{x^{2}+x-12}{x^{2}+x-30} \times \frac{x^{2}+5x+6}{x^{2}-2x-3} \div \frac{x+3}{x^{2}+7x+6}$
Factor all the expressions in numerator and in denominator if possible.
$=\frac{(x+4)(x-3)}{(x+6)(x-5)} \times \frac{(x+3)(x+2)}{(x-3)(x+1)} \div \frac{(x+3)}{(x+6)(x+1)} ; x \ne -6,-3,-1,3,5$
$ x \ne -6,-3,-1,3,5$ for non-zero denominators.
$=\frac{(x+4)(x-3)}{(x+6)(x-5)} \times \frac{(x+3)(x+2)}{(x-3)(x+1)}\times\frac{(x+6)(x+1)}{(x+3)} ; x \ne -6,-3,-1,3,5$
Divide out the common factors. The result is
$=\frac{(x+4)(x+2)}{(x-5)}; x \ne -6,-3,-1,3,5$