We have,
9xโ2<(x+2)2<12xโ3
Case I 9xโ2<x2+4x+4
\begin{gather*} x^2-5 x+6>0 \\ \Rightarrow \quad(x-3)(x-2)>0 \\ x \in(-\infty, 2) \cup(3, \infty) \,\,\,\,\,\,\,\,\,...(i) \end{gather*}
Case II x2+4x+4<12xโ3
\begin{gather*} x^2-8 x+7<0 \\ \Rightarrow \quad(x-7)(x-1)<0 \\ x \in(1,7) \,\,\,\,\,\,\,...(ii) \end{gather*}
From Eqs. (i) and (ii),
xโ(1,2)โช(3,7)
Number of integral value of x is 3 i.e., {4,5,6}.