Question

The number of integers in the domain of $f\left(x\right)=\frac{8{\log }_{10}(x^3-x)}{\sqrt{9-x^2}}$ is

Answer 1

$f\left(x\right)=\frac{8{\log }_{10}(x^3-x)}{\sqrt{9-x^2}}$
For $f\left(x\right)$ to be defined,
$9-x^2>0,x^3-x>0\Rightarrow 9>x^2\Rightarrow 3>x>-3\cdots \left(1\right)$

And
$x^3-x>0\Rightarrow x(x-1)(x+1)>0\Rightarrow x\in (-1,0)\cup (1,\infty )\cdots \left(2\right)$

Domain of $f\left(x\right)=\left(1\right)\cap \left(2\right)$
$x\in (-1,0)\cup (1,3)$
Therefore, the intergral value of $x=2$
Hence, the number of intergers in the domain is $1.$