Factor the denominator.

The denominator $x^2-16$ is a difference of squares, which can be factored as $(x-4)(x+4)$.

$$\frac{1}{x^2-16} = \frac{1}{(x-4)(x+4)}$$

For distinct linear factors, the form is:

$$\frac{1}{(x-4)(x+4)} = \frac{A}{x-4} + \frac{B}{x+4}$$

Multiply both sides by $(x-4)(x+4)$:

$$1 = A(x+4) + B(x-4)$$

To find $A$, set $x=4$:

$$1 = A(4+4) + B(4-4)$$ $$1 = 8A + 0$$ $$A = \frac{1}{8}$$

To find $B$, set $x=-4$:

$$1 = A(-4+4) + B(-4-4)$$ $$1 = 0 - 8B$$ $$B = -\frac{1}{8}$$

$$\frac{1}{x^2-16} = \frac{\frac{1}{8}}{x-4} + \frac{-\frac{1}{8}}{x+4}$$ $$\frac{1}{x^2-16} = \frac{1}{8(x-4)} - \frac{1}{8(x+4)}$$

The expression in partial fractions is:

$$\frac{1}{8(x-4)} - \frac{1}{8(x+4)}$$