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Solution
Your input: perform the partial fraction decomposition of $$$\frac{1}{r^{2} \left(r + 1\right)}$$$
The form of the partial fraction decomposition is
$$\frac{1}{r^{2} \left(r + 1\right)}=\frac{A}{r}+\frac{B}{r^{2}}+\frac{C}{r + 1}$$
Write the right-hand side as a single fraction:
$$\frac{1}{r^{2} \left(r + 1\right)}=\frac{r^{2} C + r \left(r + 1\right) A + \left(r + 1\right) B}{r^{2} \left(r + 1\right)}$$
The denominators are equal, so we require the equality of the numerators:
$$1=r^{2} C + r \left(r + 1\right) A + \left(r + 1\right) B$$
Expand the right-hand side:
$$1=r^{2} A + r^{2} C + r A + r B + B$$
Collect up the like terms:
$$1=r^{2} \left(A + C\right) + r \left(A + B\right) + B$$
The coefficients near the like terms should be equal, so the following system is obtained:
$$\begin{cases} A + C = 0\\A + B = 0\\B = 1 \end{cases}$$
Solving it (for steps, see system of equations calculator), we get that $$$A=-1$$$, $$$B=1$$$, $$$C=1$$$
Therefore,
$$\frac{1}{r^{2} \left(r + 1\right)}=\frac{-1}{r}+\frac{1}{r^{2}}+\frac{1}{r + 1}$$
Answer: $$$\frac{1}{r^{2} \left(r + 1\right)}=\frac{-1}{r}+\frac{1}{r^{2}}+\frac{1}{r + 1}$$$