部分分式分解計算器

Your input: perform the partial fraction decomposition of $$$\frac{1}{x^{4} \left(1 - x\right)^{2}}$$$

Simplify the expression: $$$\frac{1}{x^{4} \left(1 - x\right)^{2}}=\frac{1}{x^{4} \left(x - 1\right)^{2}}$$$

The form of the partial fraction decomposition is

$$\frac{1}{x^{4} \left(x - 1\right)^{2}}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x^{3}}+\frac{D}{x^{4}}+\frac{E}{x - 1}+\frac{F}{\left(x - 1\right)^{2}}$$

Write the right-hand side as a single fraction:

$$\frac{1}{x^{4} \left(x - 1\right)^{2}}=\frac{x^{4} \left(x - 1\right) E + x^{4} F + x^{3} \left(x - 1\right)^{2} A + x^{2} \left(x - 1\right)^{2} B + x \left(x - 1\right)^{2} C + \left(x - 1\right)^{2} D}{x^{4} \left(x - 1\right)^{2}}$$

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

$$1=x^{4} \left(x - 1\right) E + x^{4} F + x^{3} \left(x - 1\right)^{2} A + x^{2} \left(x - 1\right)^{2} B + x \left(x - 1\right)^{2} C + \left(x - 1\right)^{2} D$$

Expand the right-hand side:

$$1=x^{5} A + x^{5} E - 2 x^{4} A + x^{4} B - x^{4} E + x^{4} F + x^{3} A - 2 x^{3} B + x^{3} C + x^{2} B - 2 x^{2} C + x^{2} D + x C - 2 x D + D$$

Collect up the like terms:

$$1=x^{5} \left(A + E\right) + x^{4} \left(- 2 A + B - E + F\right) + x^{3} \left(A - 2 B + C\right) + x^{2} \left(B - 2 C + D\right) + x \left(C - 2 D\right) + D$$

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

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

Solving it (for steps, see system of equations calculator), we get that $$$A=4$$$, $$$B=3$$$, $$$C=2$$$, $$$D=1$$$, $$$E=-4$$$, $$$F=1$$$

Therefore,

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

Answer: $$$\frac{1}{x^{4} \left(1 - x\right)^{2}}=\frac{4}{x}+\frac{3}{x^{2}}+\frac{2}{x^{3}}+\frac{1}{x^{4}}+\frac{-4}{x - 1}+\frac{1}{\left(x - 1\right)^{2}}$$$