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Your input: perform the partial fraction decomposition of $$$\frac{1}{9 x^{2} - 4}$$$

Factor the denominator: $$$\frac{1}{9 x^{2} - 4}=\frac{1}{\left(3 x - 2\right) \left(3 x + 2\right)}$$$

The form of the partial fraction decomposition is

$$\frac{1}{\left(3 x - 2\right) \left(3 x + 2\right)}=\frac{A}{3 x - 2}+\frac{B}{3 x + 2}$$

Write the right-hand side as a single fraction:

$$\frac{1}{\left(3 x - 2\right) \left(3 x + 2\right)}=\frac{\left(3 x - 2\right) B + \left(3 x + 2\right) A}{\left(3 x - 2\right) \left(3 x + 2\right)}$$

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

$$1=\left(3 x - 2\right) B + \left(3 x + 2\right) A$$

Expand the right-hand side:

$$1=3 x A + 3 x B + 2 A - 2 B$$

Collect up the like terms:

$$1=x \left(3 A + 3 B\right) + 2 A - 2 B$$

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

$$\begin{cases} 3 A + 3 B = 0\\2 A - 2 B = 1 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=\frac{1}{4}$$$, $$$B=- \frac{1}{4}$$$

Therefore,

$$\frac{1}{\left(3 x - 2\right) \left(3 x + 2\right)}=\frac{\frac{1}{4}}{3 x - 2}+\frac{- \frac{1}{4}}{3 x + 2}$$

Answer: $$$\frac{1}{9 x^{2} - 4}=\frac{\frac{1}{4}}{3 x - 2}+\frac{- \frac{1}{4}}{3 x + 2}$$$