By definition of \(\operatorname{sech}(x)\):
\[ \int \operatorname{sech}(x) \ dx = \int \frac{2}{e^x + e^{-x}} \ dx = \int \frac{2e^x}{e^{2x} + 1} \ dx \]
Let \(u = e^x\):
\[\begin{gathered} du = e^x \ dx \\ \int \operatorname{sech}(x) \ dx = \int \frac{2}{u^2 + 1} \ du \end{gathered}\]
Since \(\frac{d}{dx} \arctan(u) = \frac{1}{u^2 + 1}\):
\[\begin{aligned} \int \operatorname{sech}(x) \ dx &= 2\int \frac{1}{u^2 + 1} \ du = 2 \arctan(u)+C \\ \int \operatorname{sech}(x) \ dx &= 2 \arctan(e^x) + C \end{aligned}\]
That is the integral of \(\operatorname{sech}(x)\). Now let's try to find the integral of \(\operatorname{csch}(x)\):
\[ \int \operatorname{csch}(x) \ dx = \int \frac{2}{e^x - e^{-x}} \ dx = \int \frac{2e^x}{e^{2x} - 1} \ dx \]
Let \(u = e^x\):
\[\begin{gathered} du = e^x \ dx \\ \int \operatorname{csch}(x) \ dx = \int \frac{2}{u^2 - 1} \ du \end{gathered}\]
Since \(\frac{d}{dx} \operatorname{artanh}(u) = \frac{1}{1 - u^2}\) for \(|u| \lt 1\) and \(\frac{d}{dx} \operatorname{arcoth}(u) = \frac{1}{1 - u^2}\) for \(|u| \gt 1\):
\[ \int \operatorname{csch}(x) \ dx = -2 \int \frac{1}{1-u^2} \ du = \begin{cases} -2 \operatorname{artanh}(u) + C &\quad \text{when }|u| \lt 1 \\ -2 \operatorname{arcoth}(u) + C &\quad \text{when }|u| \gt 1 \end{cases} \]
Since \(u=e^x\):
\[ \int \operatorname{csch}(x) \ dx = \begin{cases} -2 \operatorname{artanh}(e^x) + C &\quad \text{when }e^x<1 \text{ or } x \lt 0 \\ -2 \operatorname{arcoth}(e^x) + C &\quad \text{when } e^x >1 \text{ or } x \gt 0 \end{cases} \]