We will use integration by parts to evaluate \(\int \operatorname{arsech}(x) \ dx\). Let \(u = \operatorname{arsech}(x)\) and \(v=x\):
\[\begin{aligned} u = \operatorname{arsech}(x) &\implies du = \frac{-1}{x \sqrt{1-x^2}} \ dx, x \in (0,1) \\ v = x &\implies dv = dx \end{aligned}\]
Now let's use integration by parts:
\[ \int \operatorname{arsech}(x) \ dx = x \operatorname{arsech}(x) - \int \frac{-x}{x\sqrt{1-x^2}} = x \operatorname{arsech}(x) + \int \frac{1}{\sqrt{1-x^2}} \ dx \]
Since \(\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1-x^2}} \):
\[ \int \operatorname{arsech}(x) \ dx = x \operatorname{arsech}(x) + \arcsin(x) + C \]
Now let's try to find \(\int \operatorname{arcsch}(x) \ dx\). We will use integration by parts again:
\[\begin{aligned} u = \operatorname{arcsch}(x) &\implies du = \frac{-1}{|x| \sqrt{1+x^2}} \ dx \\ v = x &\implies dv = dx \end{aligned}\]
Now let's use integration by parts:
\[ \int \operatorname{arcsch}(x) \ dx = x \operatorname{arcsch}(x) + \int \frac{x}{|x| \sqrt{1+x^2}} \ dx \]
If \(x > 0\):
\[\begin{align} \int \operatorname{arcsch}(x) \ dx &= x \operatorname{arcsch}(x) + \int \frac{1}{\sqrt{1+x^2}} \ dx\\ &= x \operatorname{arcsch}(x) + \operatorname{arsinh}(x) + C \end{align}\]
If \(x < 0\):
\[\begin{align} \int \operatorname{arcsch}(x) \ dx &= x \operatorname{arcsch}(x) + \int \frac{-|x|}{|x|\sqrt{1+x^2}} \ dx = x \operatorname{arcsch}(x) - \int \frac{1}{\sqrt{1+x^2}} \ dx \\ &= x \operatorname{arcsch}(x) - \operatorname{arsinh}(x) + C \end{align}\]
If \(x < 0\), then \(- \operatorname{arsinh}(x) = |\operatorname{arsinh}(x)|\):
\[\int \operatorname{arcsch}(x) \ dx = x \operatorname{arcsch}(x) + |\operatorname{arsinh}(x)| + C \]
So for both \(x \lt 0\) and \(x \gt 0\):
\[\int \operatorname{arcsch}(x) \ dx = x \operatorname{arcsch}(x) + |\operatorname{arsinh}(x)| + C \]