Any smart ways for this?

$\textbf{Short Solution:}$ Just apply the rational root test.

$\textbf{Long Solution:}$ First depress the cubic by making the substitution $x=y-\frac{b}{3a}=y+\frac{24}{12}=y+2$ to get: $4y^3-25y=0$ Which can be factored as $y(4y^2-25)=y(2y+5)(2y-5)=0\rightarrow y={0,-\frac{5}{2},\frac{5}{2}}$ .So the integer solution of the original cubic is $x=0+2=\boxed{2}$