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Your input: perform the partial fraction decomposition of $$$\frac{x^{2}}{x^{4} - 2 x^{2} - 8}$$$

Factor the denominator: $$$\frac{x^{2}}{x^{4} - 2 x^{2} - 8}=\frac{x^{2}}{\left(x - 2\right) \left(x + 2\right) \left(x^{2} + 2\right)}$$$

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

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

Expand the right-hand side:

$$x^{2}=x^{3} A + x^{3} B + x^{3} C + 2 x^{2} A - 2 x^{2} B + x^{2} D + 2 x A + 2 x B - 4 x C + 4 A - 4 B - 4 D$$

Collect up the like terms:

$$x^{2}=x^{3} \left(A + B + C\right) + x^{2} \left(2 A - 2 B + D\right) + x \left(2 A + 2 B - 4 C\right) + 4 A - 4 B - 4 D$$

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

$$\begin{cases} A + B + C = 0\\2 A - 2 B + D = 1\\2 A + 2 B - 4 C = 0\\4 A - 4 B - 4 D = 0 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=\frac{1}{6}$$$, $$$B=- \frac{1}{6}$$$, $$$C=0$$$, $$$D=\frac{1}{3}$$$

Therefore,

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

Answer: $$$\frac{x^{2}}{x^{4} - 2 x^{2} - 8}=\frac{\frac{1}{6}}{x - 2}+\frac{- \frac{1}{6}}{x + 2}+\frac{\frac{1}{3}}{x^{2} + 2}$$$