Answer
$\frac{x-2}{(x+3)(x-3)}$
Work Step by Step
$\frac{1}{x+3}+\frac{1}{x^{2}-9}$
Factor $x^{2}-9$ and replace in the denominator:
$=\frac{1}{x+3}+\frac{1}{(x+3)(x-3)}$
Find the lowest common denominator (i.e. $(x+3)(x-3)$) and adjust the fractions accordingly:
$=\frac{1\times (x-3)}{(x+3)\times (x-3)}+\frac{1}{(x+3)(x-3)}$
$=\frac{x-3}{(x+3)(x-3)}+\frac{1}{(x+3)(x-3)}$
Combine the fractions:
$=\frac{x-3+1}{(x+3)(x-3)}$
Simplify:
$=\frac{x-2}{(x+3)(x-3)}$
