Simplify 1/(x+3)+1/(x^2-9)

$\frac{1}{x + 3} + \frac{1}{x^{2} - 9}$

Simplify the denominator.

Rewrite $9$ as $3^{2}$ .

$\frac{1}{x + 3} + \frac{1}{x^{2} - 3^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$\frac{1}{x + 3} + \frac{1}{(x + 3) (x - 3)}$

To write $\frac{1}{x + 3}$ as a fraction with a common denominator, multiply by $\frac{x - 3}{x - 3}$ .

$\frac{1}{x + 3} \cdot \frac{x - 3}{x - 3} + \frac{1}{(x + 3) (x - 3)}$

Multiply $\frac{1}{x + 3}$ by $\frac{x - 3}{x - 3}$ .

$\frac{x - 3}{(x + 3) (x - 3)} + \frac{1}{(x + 3) (x - 3)}$

Combine the numerators over the common denominator.

$\frac{x - 3 + 1}{(x + 3) (x - 3)}$

Add $- 3$ and $1$ .

$\frac{x - 2}{(x + 3) (x - 3)}$