部分分式分解计算器

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

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

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

Expand the right-hand side:

$$x^{2} + 1=x^{3} A + x^{3} C + x^{2} B + x^{2} D + 3 x A + 2 x C + 3 B + 2 D$$

Collect up the like terms:

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

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

$$\begin{cases} A + C = 0\\B + D = 1\\3 A + 2 C = 0\\3 B + 2 D = 1 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=0$$$, $$$B=-1$$$, $$$C=0$$$, $$$D=2$$$

Therefore,

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

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