Find the only integer solution

First, we need to find the domain for $\sqrt{2-x^4}$ which is :

$2-x^4 \geq 0$

$x^4 - 2 \leq 0$

$(x - \sqrt[4]{2})(x + \sqrt[4]{2})(x^2 + \sqrt{2}) \leq 0$

And we have the domain is $- \sqrt[4]{2} \leq x \leq \sqrt[4]{2}$ ( $x^2 + 2$ doesn't have a real root) and $x \neq 0$ because $\frac{A}{0}$ is undefined for any $A \neq 0$

We know that x is an integer, so the only possibility for x are -1 and 1

If we input x=1 we will find $\sqrt{2 - 1} < 1$ which is should be an equality ( $\sqrt{2 - 1} = 1$ )

If we input x = -1 we will get

$\frac{\sqrt{2 - 1}}{- 1} < 1$

which is true because the value of root is always positive and something positive divided by negative will result in something negative

So, the only integer that satisfy x is -1