Primitive of Inverse Hyperbolic Tangent Function

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


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

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

The result follows by setting $a = 1$.


Also presented as

This result can also be presented as:

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

Also see