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 $a = 1$.

$\blacksquare$

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

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

Sources

• 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): arc-tanh
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Appendix: Table $2$: Integrals
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Appendix: Table $2$: Integrals