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

Simplify the expression: $$$\frac{1}{u^{2} \left(1 - u^{2}\right)}=\frac{-1}{u^{4} - u^{2}}$$$

Factor the denominator: $$$\frac{-1}{u^{4} - u^{2}}=\frac{-1}{u^{2} \left(u - 1\right) \left(u + 1\right)}$$$

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

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

Expand the right-hand side:

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

Collect up the like terms:

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

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

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

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

Therefore,

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

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