Calculadora de descomposición en fracciones parciales

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

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

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

$$1=t \left(t - 1\right) B + t \left(t + 1\right) C + \left(t - 1\right) \left(t + 1\right) A$$

Expand the right-hand side:

$$1=t^{2} A + t^{2} B + t^{2} C - t B + t C - A$$

Collect up the like terms:

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

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

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

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

Therefore,

$$\frac{1}{t \left(t - 1\right) \left(t + 1\right)}=\frac{-1}{t}+\frac{\frac{1}{2}}{t + 1}+\frac{\frac{1}{2}}{t - 1}$$

Answer: $$$\frac{1}{t^{3} - t}=\frac{-1}{t}+\frac{\frac{1}{2}}{t + 1}+\frac{\frac{1}{2}}{t - 1}$$$