Question

The number of integers satisfying the inequation $\frac{x(x^2+2x+2)({\log }_{2}x-3)}{(x+3{)}^2(x^2-x-6)}\le 0,$ is

Answer 1

Clearly, $x>0\cdots \left(1\right)$
and $(x+2)(x-3)\ne 0\Rightarrow x\ne -2,3\cdots \left(2\right)$

$\frac{x(x^2+2x+2)({\log }_{2}x-3)}{(x+3{)}^2(x^2-x-6)}\le 0$
$\Rightarrow \frac{x({\log }_{2}x-3)}{(x+2)(x-3)}\le 0$
$(\because x^2+2x+2>0,(x+3{)}^2>0,\forall x>0)$

Critical points are
$x=0$
${\log }_{2}x-3\Rightarrow x=8$
$(x+2)(x-3)\Rightarrow x=-2,3$

By wavy curve method,
$x\in (-2,0]\cup (3,8]\cdots \left(3\right)$
$\left(1\right)\cap \left(2\right)\cap \left(3\right),$ we get
$x\in (3,8]$
Integers are $4,5,6,7,8$