$N=\frac{9x+5}{3x-1}=\frac{3(3x-1)+8}{3x-1}=3+\frac{8}{3x-1}$
Để N nguyên
$⇔8\vdots(3x-1)$
$⇒(3x-1)\in{Ư(8)=\{±1; ±2; ±4; ±8\}}$
$⇒x\in{\Big\{\frac{2}{3}; 0; 1; -\frac{1}{3}; \frac{5}{3}; -1; 3; -\frac{7}{3}\Big\}}$
Vì x nguyên
$⇒x\in{\{0; 1; -1; 3\}}$