We will use integration by parts to evaluate \(\int \operatorname{arsinh}(x) \ dx\). Let \(u = \operatorname{arsinh}(x)\) and \(v=x\):
\[\begin{gathered} u = \operatorname{arsinh}(x) \implies du = \frac{1}{\sqrt{x^2+1}} \ dx \\ v = x \implies dv = dx \end{gathered}\]
Keep in mind \(\sqrt{x^2+1}\) is positive. Now let's use integration by parts:
\[ \int \operatorname{arsinh}(x) \ dx = x \operatorname{arsinh}(x) - \int \frac{x}{\sqrt{x^2+1}} \ dx \]
Since \(\frac{d}{dx} \sqrt{x^2+1} = \frac{x}{\sqrt{x^2+1}}\):
\[ \int \operatorname{arsinh}(x) \ dx = x \operatorname{arsinh}(x) - \sqrt{x^2+1} + C \]
We will use integration by parts again to evaluate \(\int \operatorname{arcosh}(x) \ dx\):
\[\begin{gathered} u = \operatorname{arcosh}(x) \implies du = \frac{1}{\sqrt{x^2 - 1}} \ dx \\ v = x \implies dv = dx \end{gathered}\]
Keep in mind \(\sqrt{x^2-1}\) is positive. Now let's use integration by parts:
\[ \int \operatorname{arcosh}(x) \ dx = x \operatorname{arcosh}(x) - \int \frac{x}{\sqrt{x^2-1}} \ dx \]
Since \(\frac{d}{dx} \sqrt{x^2-1} = \frac{x}{\sqrt{x^2 -1}}\):
\[ \int \operatorname{arcosh}(x) \ dx = x \operatorname{arcosh}(x) - \sqrt{x^2-1} + C \]