[tex] \frac{3}{x + 1} \: \in \: \mathbb{Z} = > x + 1 \: \in \: D_{3}[/tex]
[tex]D_{3} = \left\{\pm1,\pm3\right\}[/tex]
[tex]x + 1 = - 1 = > x = - 2 \: \in \: \mathbb{Z}[/tex]
[tex]x + 1 = 1 = > x = 0 \: \in \: \mathbb{Z}[/tex]
[tex]x + 1 = - 3 = > x = - 4 \: \in \: \mathbb{Z}[/tex]
[tex]x + 1 = 3 = > x = 2 \: \in \: \mathbb{Z}[/tex]
[tex] - 4 + ( - 2) + 0 + 2 = - 4 - 2 + 0 + 2 = - 4[/tex]
