Hyperbolic substitution for $\int\frac{dx}{x\sqrt{1-x^2}}$ 
Substitute $x=\sinh\theta$, $\cosh\theta$ or $\tanh\theta$. After integration change back to $x$.$$\int\frac{dx}{x\sqrt{1-x^2}}$$

Substituting $x=\tanh\theta$, we have
$$\begin{align}
\int\frac{dx}{x\sqrt{1-x^2}}&=\int\frac{\text{sech}^2\,\theta\,d\theta}{\tanh\theta\sqrt{1-\tanh^2\theta}}\\
&=\int\text{csch }\theta\,d\theta\\
&=-\ln|\text{csch }\theta+\text{coth }\theta|+C
\end{align}$$
Here comes the problem. We can substitute back $\text{coth }\theta=\frac1x$ but what do we have for $\text{csch }\theta$? We have $\text{csch}^2\,\theta=\coth^2\theta-1$, but neither $\sqrt{\coth^2\theta-1}$ nor $-\sqrt{\coth^2\theta-1}$ seems to be a one-off solution for $\text{csch }\theta$. How should we proceed?

For positive $x$, another substitution that does not involve integrating $\csc{x}$ or $\operatorname{csch}{x}$ is a hyperbolic secant substitution (similar to a secant/cosecant substitution for $x^2-1$), where  $x=\operatorname{sech}{\theta}$, $dx=-\operatorname{sech}{\theta}\tanh{\theta}\,d\theta$, and $$\sqrt{1-x^2}=\sqrt{1-\operatorname{sech}^2{\theta}}=\sqrt{\tanh^2{\theta}}=\tanh{\theta}$$ since $\theta\geq0$ and $\tanh{x}$ is positive for that interval. $$\int{\frac{dx}{x\sqrt{1-x^2}}}=-\int{\frac{\operatorname{sech}{\theta}\tanh{\theta}}{\operatorname{sech}{\theta}\tanh{\theta}}d\theta}=-\int{d\theta}=-\theta+C$$ Substituting $x$ back gives $$-\operatorname{arsech}{x}+C\qquad x>0$$
For negative $x$, the substitution $x=-\operatorname{sech}{\theta}$ and $dx=\operatorname{sech}{\theta}\tanh{\theta}\,d\theta$ give $$-\operatorname{arsech}{\left(-x\right)}+C\qquad x<0$$ Combining the two branches gives $$-\operatorname{arsech}{\left|x\right|}+C$$
