부분분수 분해 계산기

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

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

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

Expand the right-hand side:

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

Collect up the like terms:

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

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

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

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

Therefore,

$$\frac{1}{u^{2} \left(u^{2} + 1\right)}=\frac{0}{u}+\frac{1}{u^{2}}+\frac{-1}{u^{2} + 1}=\frac{1}{u^{2}}+\frac{-1}{u^{2} + 1}$$

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