Difference between revisions of "Antiderivative of arctanh"
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(Created page with "==Theorem== The following formula holds: $$\displaystyle\int \mathrm{arctanh}(z) \mathrm{d}z = \dfrac{\log(1-z^2)}{2} + z \mathrm{arctanh}(z) + C,$$ where $\mathrm{arctanh}$ d...") |
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Latest revision as of 23:48, 11 December 2016
Theorem
The following formula holds: $$\displaystyle\int \mathrm{arctanh}(z) \mathrm{d}z = \dfrac{\log(1-z^2)}{2} + z \mathrm{arctanh}(z) + C,$$ where $\mathrm{arctanh}$ denotes the inverse hyperbolic tangent.
