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Your input: perform the partial fraction decomposition of $$$\frac{1}{2 x^{2} - x - 3}$$$

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

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

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

Expand the right-hand side:

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

Collect up the like terms:

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

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

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

Solving it (for steps, see system of equations calculator), we get that $$$A=- \frac{1}{5}$$$, $$$B=\frac{2}{5}$$$

Therefore,

$$\frac{1}{\left(x + 1\right) \left(2 x - 3\right)}=\frac{- \frac{1}{5}}{x + 1}+\frac{\frac{2}{5}}{2 x - 3}$$

Answer: $$$\frac{1}{2 x^{2} - x - 3}=\frac{- \frac{1}{5}}{x + 1}+\frac{\frac{2}{5}}{2 x - 3}$$$