The number of integral solutions of ${\mathrm{x}}^2+9<(\mathrm{x}+3{)}^2<8\mathrm{x}+25\text{ is }$

The given inequality is ${\mathrm{x}}^2+9<(\mathrm{x}+3{)}^2<8\mathrm{x}+25$

it implies that $x^2+9<x^2+6x+9\Rightarrow 6x>0\Rightarrow x>0$

and 

    $\begin{array}{rcl}{x}^2+6x+9& <& 8x+25\\ x^2-2x+1& <& 17\\ {\left(x-1\right)}^2& <& 17\\ x& \in & \left(1-\sqrt{17},1+\sqrt{17}\right)\end{array}$

Hence the positive integral values of $x$ are $x=1,2,3,4,5$

Therefore, the number of values of $x$ is $\boxed{5}$