部分分式分解计算器

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

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

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

Expand the right-hand side:

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

Collect up the like terms:

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

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

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

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

Therefore,

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

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