$$\log _{\left(x+\frac{7}{2}\right)}\left(\frac{x-7}{2 x-3}\right)^{2} \geq 0$$

Domain :

$$ \begin{aligned} & x+\frac{7}{2}>0 \\\\ & x>\frac{-7}{2} \\\\ & x+\frac{7}{2} \neq 1 \\\\ & x \neq \frac{-5}{2} \\\\ & \frac{x-7}{2 x-3} \neq 0 \\\\ & x \neq 7 \\\\ & x \neq \frac{3}{2} \end{aligned} $$

$$ \text { Taking intersection : } x \in\left(\frac{-7}{2}, \infty\right)-\left\{-\frac{5}{2}, \frac{3}{2}, 7\right\} $$

Now $\log _{\mathrm{a}} \mathrm{b} \geq 0$ if $\mathrm{a}>1$ and $\mathrm{b} \geq 1$

$$ \begin{gathered} \text { Or } \\\\ a \in(0,1) \text { and } b \in(0,1) \end{gathered} $$

Case I : $$ x+\frac{7}{2}>1 \text { and }\left(\frac{x-7}{2 x-3}\right)^2 \geq 1 $$

$$ \therefore $$ $$ x > -\frac{5}{2} $$

and

$$ \begin{aligned} & (2 x-3)^2-(x-7)^2 \leq 0 \\\\ & (2 x-3+x-7)(2 x-3-x+7) \leq 0 \\\\ & (3 x-10)(x+4) \leq 0 \end{aligned} $$

$$ x \in\left[-4, \frac{10}{3}\right] $$

$$ \text { Intersection : } \mathrm{x} \in\left(\frac{-5}{2}, \frac{10}{3}\right] $$

Case II : $$ x+\frac{7}{2} \in(0,1) \text { and }\left(\frac{x-7}{2 x-3}\right)^2 \in(0,1) $$

$$ \begin{aligned} & \therefore 0 < x+\frac{7}{2}<1 \\\\ & -\frac{7}{2} < x < \frac{-5}{2} \end{aligned} $$

and

$$ \begin{aligned} & \left(\frac{x-7}{2 x-3}\right)^2<1 \\\\ & \Rightarrow (x-7)^2 < (2 x-3)^2 \end{aligned} $$

$$ \therefore $$ $$ x \in(-\infty,-4) \cup\left(\frac{10}{3}, \infty\right) $$

No common values of $\mathrm{x}$.

Hence intersection with feasible region.

We get $x \in\left(\frac{-5}{2}, \frac{10}{3}\right]-\left\{\frac{3}{2}\right\}$

Integral value of $x$ are $\{-2,-1,0,1,2,3\}$

No. of integral values $=6$