The domain of the definition of the function $$f(x) = \frac{1}{4 - x^2} + \log_{10}(x^3 - x)$$ is:

We have to find all real numbers $$x$$ for which the two separate parts of the given expression

$$f(x)=\dfrac{1}{4-x^{2}}+\log_{10}\!\bigl(x^{3}-x\bigr)$$

are simultaneously well-defined.

For the rational term $$\dfrac{1}{4-x^{2}}$$ the denominator must never be zero. We write

$$4-x^{2}\neq 0$$

$$\iff\;-x^{2}\neq -4$$

$$\iff\;x^{2}\neq 4$$

$$\iff\;x\neq \pm 2.$$

There is no further restriction here because a non-zero denominator may be positive or negative; we only exclude the points where it vanishes, namely $$x=-2$$ and $$x=2.$$

For the logarithmic term $$\log_{10}\!\bigl(x^{3}-x\bigr)$$ the argument of the logarithm must be strictly positive. Recalling the basic rule

“A real logarithm $$\log_{a}(y)$$ is defined only when $$y>0,$$”

we impose

$$x^{3}-x>0.$$

We factor the cubic completely:

$$x^{3}-x = x\bigl(x^{2}-1\bigr)$$

$$=x(x-1)(x+1).$$

So we need

$$x(x-1)(x+1)>0.$$

To solve this inequality we examine the sign of the product across the critical points $$x=-1,\;0,\;1.$$ We set up intervals and test each one.

1. Interval $$(-\infty,-1):$$ choose $$x=-2.$$

$$(-2)(-2-1)(-2+1)=(-2)(-3)(-1)=6\cdot(-1)=-6<0.$$

2. Interval $$(-1,0):$$ choose $$x=-\tfrac12.$$

$$\Bigl(-\tfrac12\Bigr)\Bigl(-\tfrac12-1\Bigr)\Bigl(-\tfrac12+1\Bigr) =\Bigl(-\tfrac12\Bigr)\Bigl(-\tfrac32\Bigr)\Bigl(\tfrac12\Bigr) =\tfrac34\cdot\tfrac12=\tfrac38>0.$$

3. Interval $$(0,1):$$ choose $$x=\tfrac12.$$

$$\Bigl(\tfrac12\Bigr)\Bigl(\tfrac12-1\Bigr)\Bigl(\tfrac12+1\Bigr) =\Bigl(\tfrac12\Bigr)\Bigl(-\tfrac12\Bigr)\Bigl(\tfrac32\Bigr) =-\tfrac14\cdot\tfrac32=-\tfrac38<0.$$

4. Interval $$(1,\infty):$$ choose $$x=2.$$

$$(2)(2-1)(2+1)=2\cdot1\cdot3=6>0.$$

The product is therefore positive on

$$(-1,0)\quad\text{and}\quad(1,\infty).$$

Combining both sets of conditions:

• From the logarithm: $$x\in(-1,0)\cup(1,\infty).$$
• From the denominator: $$x\neq -2,\;2.$$ The point $$-2$$ is already outside the first set, but $$2$$ lies inside $$(1,\infty)$$ and must be removed.

Subtracting the single forbidden point $$x=2$$ from $$(1,\infty)$$ splits that interval into two pieces:

$$(1,\infty)\setminus\{2\}=(1,2)\cup(2,\infty).$$

Putting everything together, the set of all admissible $$x$$ is

$$(-1,0)\;\cup\;(1,2)\;\cup\;(2,\infty).$$

This collection exactly matches Option A.

Hence, the correct answer is Option A.