Antiderivative of arctan
From specialfunctionswiki
Theorem: The following formula holds: $$\displaystyle\int \mathrm{arctan}(z) = z\mathrm{arctan}(z) - \dfrac{1}{2}\log(1+z^2)+C,$$ where $\mathrm{arctan}$ denotes the inverse tangent and $\log$ denotes the logarithm.
Proof: █