部分分式分解計算器

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

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

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

$$-1=x B + \left(x - 1\right) A$$

Expand the right-hand side:

$$-1=x A + x B - A$$

Collect up the like terms:

$$-1=x \left(A + B\right) - A$$

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

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

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

Therefore,

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

Answer: $$$\frac{1}{- x^{2} + x}=\frac{1}{x}+\frac{-1}{x - 1}$$$