Υπολογιστής ανάλυσης σε μερικά κλάσματα

Your input: perform the partial fraction decomposition of $$$\frac{1}{\left(2 k - 3\right)^{2} \left(2 k + 1\right)^{2}}$$$

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

$$1=\left(2 k - 3\right)^{2} \left(2 k + 1\right) A + \left(2 k - 3\right)^{2} B + \left(2 k - 3\right) \left(2 k + 1\right)^{2} C + \left(2 k + 1\right)^{2} D$$

Expand the right-hand side:

$$1=8 k^{3} A + 8 k^{3} C - 20 k^{2} A + 4 k^{2} B - 4 k^{2} C + 4 k^{2} D + 6 k A - 12 k B - 10 k C + 4 k D + 9 A + 9 B - 3 C + D$$

Collect up the like terms:

$$1=k^{3} \left(8 A + 8 C\right) + k^{2} \left(- 20 A + 4 B - 4 C + 4 D\right) + k \left(6 A - 12 B - 10 C + 4 D\right) + 9 A + 9 B - 3 C + D$$

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

$$\begin{cases} 8 A + 8 C = 0\\- 20 A + 4 B - 4 C + 4 D = 0\\6 A - 12 B - 10 C + 4 D = 0\\9 A + 9 B - 3 C + D = 1 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=\frac{1}{32}$$$, $$$B=\frac{1}{16}$$$, $$$C=- \frac{1}{32}$$$, $$$D=\frac{1}{16}$$$

Therefore,

$$\frac{1}{\left(2 k - 3\right)^{2} \left(2 k + 1\right)^{2}}=\frac{\frac{1}{32}}{2 k + 1}+\frac{\frac{1}{16}}{\left(2 k + 1\right)^{2}}+\frac{- \frac{1}{32}}{2 k - 3}+\frac{\frac{1}{16}}{\left(2 k - 3\right)^{2}}$$

Answer: $$$\frac{1}{\left(2 k - 3\right)^{2} \left(2 k + 1\right)^{2}}=\frac{\frac{1}{32}}{2 k + 1}+\frac{\frac{1}{16}}{\left(2 k + 1\right)^{2}}+\frac{- \frac{1}{32}}{2 k - 3}+\frac{\frac{1}{16}}{\left(2 k - 3\right)^{2}}$$$