Partial Fraction Decomposition Calculator

Your input: perform the partial fraction decomposition of $$$\frac{1}{6 x^{3} - 7 x^{2} - 3 x}$$$

Simplify the expression: $$$\frac{1}{6 x^{3} - 7 x^{2} - 3 x}=\frac{1}{x \left(6 x^{2} - 7 x - 3\right)}$$$

Factor the denominator: $$$\frac{1}{x \left(6 x^{2} - 7 x - 3\right)}=\frac{1}{x \left(2 x - 3\right) \left(3 x + 1\right)}$$$

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

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

Expand the right-hand side:

$$1=6 x^{2} A + 2 x^{2} B + 3 x^{2} C - 7 x A - 3 x B + x C - 3 A$$

Collect up the like terms:

$$1=x^{2} \left(6 A + 2 B + 3 C\right) + x \left(- 7 A - 3 B + C\right) - 3 A$$

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

$$\begin{cases} 6 A + 2 B + 3 C = 0\\- 7 A - 3 B + C = 0\\- 3 A = 1 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=- \frac{1}{3}$$$, $$$B=\frac{9}{11}$$$, $$$C=\frac{4}{33}$$$

Therefore,

$$\frac{1}{x \left(2 x - 3\right) \left(3 x + 1\right)}=\frac{- \frac{1}{3}}{x}+\frac{\frac{9}{11}}{3 x + 1}+\frac{\frac{4}{33}}{2 x - 3}$$

Answer: $$$\frac{1}{6 x^{3} - 7 x^{2} - 3 x}=\frac{- \frac{1}{3}}{x}+\frac{\frac{9}{11}}{3 x + 1}+\frac{\frac{4}{33}}{2 x - 3}$$$