Difference between revisions of "Antiderivative of arctan"

Revision as of 07:26, 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.

Proof

References