Answer
$D=(-\infty,-2) \cup(-1, 1) \cup (2, \infty) $
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a Rational function.
$\displaystyle f(x)= \frac{1}{\sqrt{x^4-5x^2+4}}$,
Lets take the denominator of $f(x)$ and factorise and assign it to $g(x)$.
$\sqrt {x^4-5x^2+4} \gt 0,$
$x^4-5x^2+4 \gt 0,$ lets let $x^2=k$.
$k^2-5k+4 \gt 0,$
$k^2-k-4k+4 \gt 0,$
$(k-1)(k-4),$ substituting back in $x^2=k$.
$(x^2-1)(x^2-4)\gt0,$
$g(x)=(x-1)(x+1)(x-2)(x+2)$
2.The cut points are:
$\displaystyle (x-1)(x+1)(x-2)(x+2)=0$
$x=-2$ or $x=-1$ or $x=1$ or $x=2$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $g(x)$ and eventually $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & g(a),signs & f(a), signs\\
& & \displaystyle (a-1)(a+1)(a-2)(a+2) & \displaystyle \frac{1}{\sqrt{(a-1)(a+1)(a-2)(a+2)}}& \\
(-\infty,-2) & -5 & (-)(-)(-)(-)=(+)&\frac{}{(-)(-)(-)(-)}=(+) \\
(-2, -1) &\frac{3}{2}& (-)(-)(-)(+)=(-) & \frac{}{(-)(-)(-)(+)}=Undefined \\
(-1, 1) & 0 & (-)(+)(-)(+)=(+)&\frac{}{(-)(+)(-)(+)}=(+)\\
(1, 2) & \frac{3}{2} & (+)(+)(-)(+)=(-)&\frac{}{(+)(+)(-)(+)}=Undefined \\
(2,\infty) & 5 & (+)(+)(+)(+)=(+)&\frac{}{(+)(+)(+)(+)}=(+) \\
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
The Domain of $f(x)$ is: $D=(-\infty,-2) \cup(-1, 1) \cup (2, \infty) $
