Answer
$\frac{2x+7}{(x+3)(x+4)}$
Work Step by Step
We form a common denominator, then add, factor, and simplify:
$\displaystyle \frac{2}{x+3}-\frac{1}{x^{2}+7x+12}=\frac{2}{x+3}-\frac{1}{(x+3)(x+4)}=\frac{2(x+4)}{(x+3)(x+4)}+\frac{-1}{(x+3)(x+4)}
=\frac{2x+8-1}{(x+3)(x+4)}=\frac{2x+7}{(x+3)(x+4)}$
