Primitive of Inverse Hyperbolic Cotangent Function

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Theorem

$\ds \int \arcoth x \rd x = x \arcoth x + \frac {\map \ln {x^2 - 1} } 2 + C$


Proof

From Primitive of $\arcoth \dfrac x a$:

$\ds \int \arcoth \frac x a \rd x = x \arcoth \dfrac x a + \frac {a \map \ln {x^2 - a^2} } 2 + C$

The result follows by setting $a = 1$.

$\blacksquare$


Also presented as

This result can also be presented as:

$\ds \int \arcoth x \rd x = x \arcoth x + \ln \sqrt {x^2 - 1} + C$


Also see


Sources