Lời giải:
\(\frac{x^2+2x}{x+1}=\frac{x(x+1)+(x+1)-1}{x+1}=\frac{(x+1)(x+1)-1}{x+1}=x+1-\frac{1}{x+1}\)
Để với $x\in\mathbb{Z}$, $\frac{x^2+2x}{x+1}\in\mathbb{Z}$ thì:
\(x+1-\frac{1}{x+1}\in\mathbb{Z}\Leftrightarrow \frac{1}{x+1}\in\mathbb{Z}\Leftrightarrow 1\vdots x+1\)
\(\Rightarrow x+1\in\left\{\pm 1\right\}\Rightarrow x\in\left\{0;-2\right\}\)