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Solution
Your input: perform the partial fraction decomposition of $$$\frac{1}{100 - x^{2}}$$$
Simplify the expression: $$$\frac{1}{100 - x^{2}}=\frac{-1}{x^{2} - 100}$$$
Factor the denominator: $$$\frac{-1}{x^{2} - 100}=\frac{-1}{\left(x - 10\right) \left(x + 10\right)}$$$
The form of the partial fraction decomposition is
$$\frac{-1}{\left(x - 10\right) \left(x + 10\right)}=\frac{A}{x - 10}+\frac{B}{x + 10}$$
Write the right-hand side as a single fraction:
$$\frac{-1}{\left(x - 10\right) \left(x + 10\right)}=\frac{\left(x - 10\right) B + \left(x + 10\right) A}{\left(x - 10\right) \left(x + 10\right)}$$
The denominators are equal, so we require the equality of the numerators:
$$-1=\left(x - 10\right) B + \left(x + 10\right) A$$
Expand the right-hand side:
$$-1=x A + x B + 10 A - 10 B$$
Collect up the like terms:
$$-1=x \left(A + B\right) + 10 A - 10 B$$
The coefficients near the like terms should be equal, so the following system is obtained:
$$\begin{cases} A + B = 0\\10 A - 10 B = -1 \end{cases}$$
Solving it (for steps, see system of equations calculator), we get that $$$A=- \frac{1}{20}$$$, $$$B=\frac{1}{20}$$$
Therefore,
$$\frac{-1}{\left(x - 10\right) \left(x + 10\right)}=\frac{- \frac{1}{20}}{x - 10}+\frac{\frac{1}{20}}{x + 10}$$
Answer: $$$\frac{1}{100 - x^{2}}=\frac{- \frac{1}{20}}{x - 10}+\frac{\frac{1}{20}}{x + 10}$$$