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

The form of the partial fraction decomposition is

$$\frac{1}{k \left(k + 2\right)}=\frac{A}{k}+\frac{B}{k + 2}$$

Write the right-hand side as a single fraction:

$$\frac{1}{k \left(k + 2\right)}=\frac{k B + \left(k + 2\right) A}{k \left(k + 2\right)}$$

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

$$1=k B + \left(k + 2\right) A$$

Expand the right-hand side:

$$1=k A + k B + 2 A$$

Collect up the like terms:

$$1=k \left(A + B\right) + 2 A$$

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

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

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

Therefore,

$$\frac{1}{k \left(k + 2\right)}=\frac{\frac{1}{2}}{k}+\frac{- \frac{1}{2}}{k + 2}$$

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