Calculus Beyond the imagination.

Let $g(x) = (2^{\pi/\tan^{-1}x}-4)(x-4)(x-10)$ , $h(x) = x!-(x-1)!$ and $f(x) = \dfrac {g(x)}{h(x)}$ .

Considering $h(x)$ : We note that $n!$ , where $n$ is an integer, is real if $n \ge 0$ , therefore, $x-1\ge 0 \implies x \ge 1$ . But when $x = 1$ , $\implies h(x) = x! - (x-1)! = 1! - 0! = 1-1 = 0$ , $\implies f(x) \to \infty$ , which is undefined. Therefore, $x \ge 2$ . and for $x \ge 2$ , $h(x) = x! - (x-1)! > 0$ , $\implies f(x) < 0$ when $g(x)< 0$ .

Considering $g(x)$ : We note that $g(x) < 0$ , when one or all three of its factors $g_1(x) = 2^{\pi/\tan^{-1}x}-4$ , $g_2(x) = x-4$ and $g_3(x) = x-10$ are negative.

• Considering $g_1(x)$ : The minimum value of $\min (2^{\pi/\tan^{-1} x}) = 2^2 = 4$ , when $x \to \infty$ . Therefore, $g_1(x) = 2^{\pi/\tan^{-1}x}-4 > 0$ for $x > 0$ .
• And $g(x) < 0$ when $g_2(x) = x - 4 > 0 \implies x \ge 5$ and $g_3(x) = x - 10 < 0 \implies x \le 9$ .

Therefore, $f(x) < 0$ , when $5 \le x \le 9$ , and $x$ has $\boxed{5}$ integral values.