[tex]\it E(X)=\dfrac{X^2+8X+15}{X^2+5X}+\dfrac{X-2}{X} =\dfrac{X^2+5X+3X+15}{X(X+5)}+ \dfrac{X-2}{X}=\\ \\ \\ =\dfrac{X(X+5)+3(X+5)}{X(X+5)}+\dfrac{X-2}{X}=\dfrac{(X+5)(X+3)}{X(X+5)}+\dfrac{X-2}{X}=\\ \\ \\ = \dfrac{X+3}{X}+\dfrac{X-2}{X}=\dfrac{X+3+X-2}{X}=\dfrac{2X+1}{X}[/tex]
[tex]\it E(X)\in\mathbb{Z} \Rightarrow \dfrac{2X+1}{X}\in\mathbb{Z}\Rightarrow \dfrac{2X}{X}+\dfrac{1}{X}\in\mathbb{Z}\Rightarrow 2+\dfrac{1}{X}\in\mathbb{Z}\Rightarrow\\ \\ \\ \Rightarrow\dfrac{1}{X}\in\mathbb{Z} \Rightarrow X\in\{-1,\ \ 1\}[/tex]
