Given :
$f(x) = 2x + 1$
$g(x) = x^2 - 2$
Now,
$$(f \circ g)(x) = (g \circ f)(x)$$
$$\Rightarrow f(g(x)) = g(f(x)) $$
$$\Rightarrow f(x^2 - 2)= g(2x + 1)$$
$$\Rightarrow 2(x^2 - 2) + 1= (2x + 1)^2 - 2$$
$$\Rightarrow 2x^2 - 3 = 4x^2 + 4x - 1$$
$$\Rightarrow 0 = 4x^2 - 2x^2 + 4x - 1 + 3$$
$$\Rightarrow 0 = 2x^2 + 4x + 2$$
$$\Rightarrow 0 = x^2 + 2x + 1$$
$$\Rightarrow 0 = (x + 1)^2$$
$$\Rightarrow x + 1 = 0$$
$$\Rightarrow x = -1$$
Since $-1$ is an integer, it satisfies the condition in the problem.
Answer: The integer value of $x$ is $\boxed{\mathbf{-1}}$.
