部分分式分解計算器

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

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

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

Expand the right-hand side:

$$1=x^{3} A + x^{3} C - 5 x^{2} A + x^{2} B - 4 x^{2} C + x^{2} D + 8 x A - 4 x B + 5 x C - 2 x D - 4 A + 4 B - 2 C + D$$

Collect up the like terms:

$$1=x^{3} \left(A + C\right) + x^{2} \left(- 5 A + B - 4 C + D\right) + x \left(8 A - 4 B + 5 C - 2 D\right) - 4 A + 4 B - 2 C + D$$

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

$$\begin{cases} A + C = 0\\- 5 A + B - 4 C + D = 0\\8 A - 4 B + 5 C - 2 D = 0\\- 4 A + 4 B - 2 C + D = 1 \end{cases}$$

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

Therefore,

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

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