部分分数分解計算機

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

Factor the denominator: $$$\frac{x^{2}}{\left(2 x^{2} - 1\right)^{2}}=\frac{x^{2}}{4 \left(x - \frac{\sqrt{2}}{2}\right)^{2} \left(x + \frac{\sqrt{2}}{2}\right)^{2}}$$$

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

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

Expand the right-hand side:

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

Collect up the like terms:

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

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

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

Solving it (for steps, see system of equations calculator), we get that $$$A=- \frac{\sqrt{2}}{16}$$$, $$$B=\frac{1}{16}$$$, $$$C=\frac{\sqrt{2}}{16}$$$, $$$D=\frac{1}{16}$$$

Therefore,

$$\frac{\frac{x^{2}}{4}}{\left(x - \frac{\sqrt{2}}{2}\right)^{2} \left(x + \frac{\sqrt{2}}{2}\right)^{2}}=\frac{- \frac{\sqrt{2}}{8}}{2 x + \sqrt{2}}+\frac{\frac{1}{4}}{\left(2 x + \sqrt{2}\right)^{2}}+\frac{\frac{\sqrt{2}}{8}}{2 x - \sqrt{2}}+\frac{\frac{1}{4}}{\left(2 x - \sqrt{2}\right)^{2}}$$

Answer: $$$\frac{x^{2}}{\left(2 x^{2} - 1\right)^{2}}=\frac{- \frac{\sqrt{2}}{8}}{2 x + \sqrt{2}}+\frac{\frac{1}{4}}{\left(2 x + \sqrt{2}\right)^{2}}+\frac{\frac{\sqrt{2}}{8}}{2 x - \sqrt{2}}+\frac{\frac{1}{4}}{\left(2 x - \sqrt{2}\right)^{2}}$$$