Partial Fraction Decomposition Calculator

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

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

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

$$1=\left(x - 1\right) \left(x + 1\right) \left(B x + C\right) + \left(x - 1\right) \left(x^{2} + 1\right) A + \left(x + 1\right) \left(x^{2} + 1\right) D$$

Expand the right-hand side:

$$1=x^{3} A + x^{3} B + x^{3} D - x^{2} A + x^{2} C + x^{2} D + x A - x B + x D - A - C + D$$

Collect up the like terms:

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

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

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

Solving it (for steps, see system of equations calculator), we get that $$$A=- \frac{1}{4}$$$, $$$B=0$$$, $$$C=- \frac{1}{2}$$$, $$$D=\frac{1}{4}$$$

Therefore,

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

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