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Your input: perform the partial fraction decomposition of $$$\frac{1}{u^{2} - 9}$$$

Factor the denominator: $$$\frac{1}{u^{2} - 9}=\frac{1}{\left(u - 3\right) \left(u + 3\right)}$$$

The form of the partial fraction decomposition is

$$\frac{1}{\left(u - 3\right) \left(u + 3\right)}=\frac{A}{u - 3}+\frac{B}{u + 3}$$

Write the right-hand side as a single fraction:

$$\frac{1}{\left(u - 3\right) \left(u + 3\right)}=\frac{\left(u - 3\right) B + \left(u + 3\right) A}{\left(u - 3\right) \left(u + 3\right)}$$

The denominators are equal, so we require the equality of the numerators:

$$1=\left(u - 3\right) B + \left(u + 3\right) A$$

Expand the right-hand side:

$$1=u A + u B + 3 A - 3 B$$

Collect up the like terms:

$$1=u \left(A + B\right) + 3 A - 3 B$$

The coefficients near the like terms should be equal, so the following system is obtained:

$$\begin{cases} A + B = 0\\3 A - 3 B = 1 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=\frac{1}{6}$$$, $$$B=- \frac{1}{6}$$$

Therefore,

$$\frac{1}{\left(u - 3\right) \left(u + 3\right)}=\frac{\frac{1}{6}}{u - 3}+\frac{- \frac{1}{6}}{u + 3}$$

Answer: $$$\frac{1}{u^{2} - 9}=\frac{\frac{1}{6}}{u - 3}+\frac{- \frac{1}{6}}{u + 3}$$$