部分分式分解计算器

Your input: perform the partial fraction decomposition of $$$\frac{1}{x^{2} - 16}$$$

Factor the denominator: $$$\frac{1}{x^{2} - 16}=\frac{1}{\left(x - 4\right) \left(x + 4\right)}$$$

The form of the partial fraction decomposition is

$$\frac{1}{\left(x - 4\right) \left(x + 4\right)}=\frac{A}{x - 4}+\frac{B}{x + 4}$$

Write the right-hand side as a single fraction:

$$\frac{1}{\left(x - 4\right) \left(x + 4\right)}=\frac{\left(x - 4\right) B + \left(x + 4\right) A}{\left(x - 4\right) \left(x + 4\right)}$$

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

$$1=\left(x - 4\right) B + \left(x + 4\right) A$$

Expand the right-hand side:

$$1=x A + x B + 4 A - 4 B$$

Collect up the like terms:

$$1=x \left(A + B\right) + 4 A - 4 B$$

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

$$\begin{cases} A + B = 0\\4 A - 4 B = 1 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=\frac{1}{8}$$$, $$$B=- \frac{1}{8}$$$

Therefore,

$$\frac{1}{\left(x - 4\right) \left(x + 4\right)}=\frac{\frac{1}{8}}{x - 4}+\frac{- \frac{1}{8}}{x + 4}$$

Answer: $$$\frac{1}{x^{2} - 16}=\frac{\frac{1}{8}}{x - 4}+\frac{- \frac{1}{8}}{x + 4}$$$