Answer
$\frac{x+5}{(2x+3)(x+4)}$
Work Step by Step
We switch from division to multiplication by taking the reciprocal of the right fraction. Then we factor and simplify:
$\displaystyle \frac{x+3}{4x^{2}-9}\div\frac{x^{2}+7x+12}{2x^{2}+7x-15}=\frac{x+3}{4x^{2}-9}*\frac{2x^{2}+7x-15}{x^{2}+7x+12}=\frac{x+3}{(2x-3)(2x+3)}*\frac{(x+5)(2x-3)}{(x+3)(x+4)}=\frac{x+5}{(2x+3)(x+4)}$
