\(A=\frac{n^2+2n}{n+3}=\frac{n^2+3n-n-3+3}{n+3}=\frac{n\left(n+3\right)-\left(n+3\right)+3}{n+3}=n-1+\frac{3}{n+3}\)
Để A là số nguyên thì \(\frac{3}{n+3}\) là số nguyên
\(\Rightarrow3⋮n+3\)
\(\Leftrightarrow n+3\inƯ\left(3\right)\)
\(\Leftrightarrow n+3\in\left\{1;3;-1;-3\right\}\)
\(\Leftrightarrow n\in\left\{-2;0;-4;-6\right\}\)
Vậy \(n\in\left\{-2;0;-4;-6\right\}\)