We will use integration by parts to evaluate \(\int \operatorname{artanh}(x) \ dx\). Let \(u = \operatorname{artanh}(x)\) and \(v=x\):
\[\begin{gathered} u = \operatorname{artanh}(x) \implies du = \frac{1}{1-x^2} \ dx, \text{ where } |x| \lt 1 \\ v = x \implies dv = dx \end{gathered}\]
Now let's use integration by parts:
\[ \int \operatorname{artanh}(x) \ dx = x \operatorname{artanh}(x) - \int \frac{x}{1-x^2} \ dx \]
Since \(\frac{d}{dx} \ln(1-x^2) = \frac{1}{1-x^2} * -2x\):
\[ \int \operatorname{artanh}(x) \ dx = x \operatorname{artanh}(x) - \left( \frac{1}{-2} \ln |1-x^2| \right) + C \]
And this leads to our answer:
\[ \int \operatorname{artanh}(x) \ dx = x \operatorname{artanh}(x) + \frac{1}{2} \ln |1-x^2| + C\]
Keep in mind that in the integral of \(\operatorname{artanh}(x)\), \(|x| \lt 1\). Now we will use integration by parts again to evaluate \(\int \operatorname{arcoth}(x) \ dx\):
\[\begin{gathered} u = \operatorname{arcoth}(x) \implies du = \frac{1}{1 - x^2} \ dx, \text{ where } |x| \gt 1 \\ v = x \implies dv = dx \end{gathered}\]
Now let's use integration by parts:
\[ \int \operatorname{arcoth}(x) \ dx = x \operatorname{arcoth}(x) - \int \frac{x}{x^2-1} \ dx \]
And this leads to our final answer:
\[ \int \operatorname{arcoth}(x) \ dx = x \operatorname{arcoth}(x) + \frac{1}{2} \ln |1-x^2| + C\]
Keep in mind that in the integral of \(\operatorname{arcoth}(x)\), \(|x| \gt 1\).