The number of integral solution $$x$$ of $$\log_{x + \frac{7}{2}}\left(\frac{x-7}{2x-3}\right)^2 \geq 0$$ is

For the logarithm to be defined,

$$x+\frac72>0,$$

$$x+\frac72\ne1,$$

and

$$\left(\frac{x-7}{2x-3}\right)^2>0.$$

Thus,

$$x>-\frac72,\qquad x\ne-\frac52,\qquad x\ne\frac32,\qquad x\ne7.$$

Since $$x$$ is an integer,

$$x\ge-3,$$

with

$$x\ne-2,\ 7.$$

Now consider the base.

Case 1:

$$x+\frac72>1$$

$$x>-\frac52.$$

For an integer,

$$x\ge-2.$$

Since the base is greater than $$1$$,

$$\log_aN\ge0\iff N\ge1.$$

Hence,

$$\left(\frac{x-7}{2x-3}\right)^2\ge1.$$

This gives

$$|x-7|\ge|2x-3|.$$

Squaring,

$$(x-7)^2\ge(2x-3)^2.$$

$$x^2-14x+49\ge4x^2-12x+9.$$

$$3x^2+2x-40\le0.$$

$$(3x-10)(x+4)\le0.$$

Hence,

$$-4\le x\le\frac{10}{3}.$$

Since

$$x\ge-2,$$

the integral values are

$$-1,\ 0,\ 1,\ 2,\ 3.$$

(Note that $$x=-2$$ is excluded because the base becomes $$\frac32$$? Actually $$x=-2$$ makes the base $$\frac32$$ but the argument is defined; however, $$x=-2$$ was excluded earlier since $$x+\frac72=1.5$$? The excluded value is $$x=-\frac52$$, so $$x=-2$$ is allowed.)

Thus the integral values are

$$-2,\ -1,\ 0,\ 1,\ 2,\ 3.$$

Case 2:

$$0<x+\frac72<1.$$

This requires

$$-\frac72<x<-\frac52,$$

which contains no integer.

Hence there are no solutions from this case.

Therefore, the total number of integral solutions is

$$6.$$

Hence,

$$\boxed{6}$$