Question: I thought that $\theta$ must be chosen such that $\cosh(\theta)$ has a range that is equal to the domain of $\frac{1}{(x^2-1)^{3/2}}$. But this can't be done since the domain includes negative numbers, and $\cosh(\theta)$ is always positive. If this substitution is made, the answer seems to be valid for all values of x, why does it work?

Answer: 
If you look for a primitive in $(1,+\infty)$,

put $$x=\cosh(\theta)$$

with $$dx=\sinh(\theta) d\theta$$

but if you want a primitive in $(-\infty,-1)$, you should put
$$x=-\cosh(\theta)$$

with $$dx=-\sinh(\theta)d\theta$$
$$I=\int\left(x^2-1\right)^{-3/2}dx$$
we want this term $x^2-1$ to be simplified to a single term, so think of trig identities.:
$$\cos^2x+\sin^2x\equiv1,\,\cosh^2x-\sinh^2x\equiv1$$
are the ones you need to know. since it is of the form $x^2-b^2$ rather than $b^2-x^2$ or $x^2+b^2$ we know to use hyperbolic functions. Manipulating the right rule, we obtain:
$$\sinh^2x=\cosh^2x-1$$
now we can use this. let:
$x=\cosh(t)$ and we get $dx=\sinh(t)$ and rewrite the integral as:
$$I=\int\left(\cosh^2(t)-1\right)^{-3/2}.\sinh(t)dt=\int(sinh(t))^{-3}.\sinh(t)dt=\int\text{csch}^2(t)dt=-\coth(t)+C$$