ta có:
$\frac{3n+1}{n+2}$=$\frac{3n+6-5}{n+2}$=$\frac{3n+6}{n+2}$+$\frac{-5}{n+2}$=3+$\frac{-5}{n+2}$
để 3n+1 chia hết cho n+2
⇔-5 phải chia hết cho n+2
⇔n+2∈Ư(-5)={±1;±5}
$+)n+2=1⇔n=-1$
$+)n+2=-1⇔n=-3$
$+)n+2=5⇔n=3$
$+)n+2=-5⇔n=-7$
vậy$ n={-1;-3;3;-7}$