\(A=\frac{\left(x^2-x-2\right)+5}{x+1}=\frac{\left(x+1\right)\left(x-2\right)+5}{x+1}=x-2+\frac{5}{x+1}\)
\(A\) nguyên \(\Leftrightarrow5⋮x+1\) hay \(x+1=Ư\left(5\right)\)
\(\Rightarrow x+1=\left\{-5;-1;1;5\right\}\)
\(\Rightarrow x=\left\{-6;-2;0;4\right\}\)