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Your input: perform the partial fraction decomposition of $$$\frac{x^{2} + 1}{x \left(x^{2} - 1\right)}$$$

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

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

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

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

Expand the right-hand side:

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

Collect up the like terms:

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

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

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

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

Therefore,

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

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