Difference between revisions of "Antiderivative of arctan"
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Revision as of 07:27, 8 June 2016
Theorem
The following formula holds: $$\displaystyle\int \mathrm{arctan}(z) \mathrm{d}z = z\mathrm{arctan}(z) - \dfrac{1}{2}\log \left(1+z^2 \right)+C,$$ where $\mathrm{arctan}$ denotes the inverse tangent and $\log$ denotes the logarithm.
