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

Simplify the expression: $\frac{1}{1 - x^{2}}=\frac{-1}{x^{2} - 1}$

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

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

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

Expand the right-hand side:

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

Collect up the like terms:

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

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

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

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

Therefore,

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

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