Partial Fraction Decomposition Calculator

Your input: perform the partial fraction decomposition of $$$\frac{x^{2} - 3}{x^{3} - 72 x}$$$

Simplify the expression: $$$\frac{x^{2} - 3}{x^{3} - 72 x}=\frac{x^{2} - 3}{x \left(x^{2} - 72\right)}$$$

Factor the denominator: $$$\frac{x^{2} - 3}{x \left(x^{2} - 72\right)}=\frac{x^{2} - 3}{x \left(x - 6 \sqrt{2}\right) \left(x + 6 \sqrt{2}\right)}$$$

The form of the partial fraction decomposition is

$$\frac{x^{2} - 3}{x \left(x - 6 \sqrt{2}\right) \left(x + 6 \sqrt{2}\right)}=\frac{A}{x}+\frac{B}{x - 6 \sqrt{2}}+\frac{C}{x + 6 \sqrt{2}}$$

Write the right-hand side as a single fraction:

$$\frac{x^{2} - 3}{x \left(x - 6 \sqrt{2}\right) \left(x + 6 \sqrt{2}\right)}=\frac{x \left(x - 6 \sqrt{2}\right) C + x \left(x + 6 \sqrt{2}\right) B + \left(x - 6 \sqrt{2}\right) \left(x + 6 \sqrt{2}\right) A}{x \left(x - 6 \sqrt{2}\right) \left(x + 6 \sqrt{2}\right)}$$

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

$$x^{2} - 3=x \left(x - 6 \sqrt{2}\right) C + x \left(x + 6 \sqrt{2}\right) B + \left(x - 6 \sqrt{2}\right) \left(x + 6 \sqrt{2}\right) A$$

Expand the right-hand side:

$$x^{2} - 3=x^{2} A + x^{2} B + x^{2} C + 6 \sqrt{2} x B - 6 \sqrt{2} x C - 72 A$$

Collect up the like terms:

$$x^{2} - 3=x^{2} \left(A + B + C\right) + x \left(6 \sqrt{2} B - 6 \sqrt{2} C\right) - 72 A$$

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

$$\begin{cases} A + B + C = 1\\6 \sqrt{2} B - 6 \sqrt{2} C = 0\\- 72 A = -3 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=\frac{1}{24}$$$, $$$B=\frac{23}{48}$$$, $$$C=\frac{23}{48}$$$

Therefore,

$$\frac{x^{2} - 3}{x \left(x - 6 \sqrt{2}\right) \left(x + 6 \sqrt{2}\right)}=\frac{\frac{1}{24}}{x}+\frac{\frac{23}{48}}{x - 6 \sqrt{2}}+\frac{\frac{23}{48}}{x + 6 \sqrt{2}}$$

Answer: $$$\frac{x^{2} - 3}{x^{3} - 72 x}=\frac{\frac{1}{24}}{x}+\frac{\frac{23}{48}}{x - 6 \sqrt{2}}+\frac{\frac{23}{48}}{x + 6 \sqrt{2}}$$$