[tex]\dfrac{2x+9}{x+1} \in \mathbb_{Z}$ $ \\ \\ \dfrac{2x+9}{x+1} = \dfrac{2x+2+7}{x+1} = \dfrac{2(x+1)+7}{x+1} = \dfrac{2(x+1)}{x+1}+\dfrac{7}{x+1} =\\ \\ = \underset{\in\mathbb_{Z}}{2}+\dfrac{7}{x+1} $ \\ \\\\ \dfrac{7}{x+1}\in \mathbb _{Z}\Rightarrow $ $ (x+1)$ $|$ $7 \Rightarrow (x+1)\in D_7 \Rightarrow (x+1)\in \Big\{\pm 1,\pm 7\Big\} \Rightarrow \\ \\ \Rightarrow (x+1)\in \Big\{-7,-1,1,7\Big\}\Big|-1 \Rightarrow \boxed{x\in \Big\{-8,-2,0,6\Big\}}[/tex]
