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Your input: perform the partial fraction decomposition of $$$\frac{x + 7}{x^{2} + 3 x + 2}$$$

Simplify the expression: $$$\frac{x + 7}{x^{2} + 3 x + 2}=\frac{x + 7}{\left(x + 1\right) \left(x + 2\right)}$$$

The form of the partial fraction decomposition is

$$\frac{x + 7}{\left(x + 1\right) \left(x + 2\right)}=\frac{A}{x + 1}+\frac{B}{x + 2}$$

Write the right-hand side as a single fraction:

$$\frac{x + 7}{\left(x + 1\right) \left(x + 2\right)}=\frac{\left(x + 1\right) B + \left(x + 2\right) A}{\left(x + 1\right) \left(x + 2\right)}$$

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

$$x + 7=\left(x + 1\right) B + \left(x + 2\right) A$$

Expand the right-hand side:

$$x + 7=x A + x B + 2 A + B$$

Collect up the like terms:

$$x + 7=x \left(A + B\right) + 2 A + B$$

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

$$\begin{cases} A + B = 1\\2 A + B = 7 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=6$$$, $$$B=-5$$$

Therefore,

$$\frac{x + 7}{\left(x + 1\right) \left(x + 2\right)}=\frac{6}{x + 1}+\frac{-5}{x + 2}$$

Answer: $$$\frac{x + 7}{x^{2} + 3 x + 2}=\frac{6}{x + 1}+\frac{-5}{x + 2}$$$