Partial Fraction Decomposition Calculator

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

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

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

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

Expand the right-hand side:

$$1=x A + x B - \sqrt{3} A + \sqrt{3} B$$

Collect up the like terms:

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

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

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

Solving it (for steps, see system of equations calculator), we get that $$$A=- \frac{\sqrt{3}}{6}$$$, $$$B=\frac{\sqrt{3}}{6}$$$

Therefore,

$$\frac{1}{\left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right)}=\frac{- \frac{\sqrt{3}}{6}}{x + \sqrt{3}}+\frac{\frac{\sqrt{3}}{6}}{x - \sqrt{3}}$$

Answer: $$$\frac{1}{x^{2} - 3}=\frac{- \frac{\sqrt{3}}{6}}{x + \sqrt{3}}+\frac{\frac{\sqrt{3}}{6}}{x - \sqrt{3}}$$$