Primitive of Inverse Hyperbolic Tangent Function

Theorem

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

Proof
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 see

 * Primitive of $\arsinh x$
 * Primitive of $\arcosh x$
 * Primitive of $\arcoth x$