[tex]\it \dfrac{3x+2}{x+1} = \dfrac{3x+3-1}{x+1} = \dfrac{3x+3}{x+1} -\dfrac{1}{x+1} =\dfrac{3(x+1)}{x+1}-\dfrac{1}{x+1} =
\\ \\ \\
= 3 -\dfrac{1}{x+1} \in \mathbb{Z} \Rightarrow x+1\in D_1 \Rightarrow x+1\in \{-1,\ 1\}|_{-1} \Rightarrow
\\ \\ \\
\Rightarrow x\in \{-2,\ 0\} \Rightarrow A = \{-2,\ 0\}[/tex]
