Difference between revisions of "Antiderivative of arctan"
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<strong>[[Antiderivative of arctan|Theorem]]:</strong> The following formula holds: | <strong>[[Antiderivative of arctan|Theorem]]:</strong> The following formula holds: | ||
| − | $$\displaystyle\int \mathrm{arctan}(z) = z\mathrm{arctan}(z) - \dfrac{1}{2}\log(1+z^2)+C,$$ | + | $$\displaystyle\int \mathrm{arctan}(z) \mathrm{d}z = z\mathrm{arctan}(z) - \dfrac{1}{2}\log(1+z^2)+C,$$ |
where $\mathrm{arctan}$ denotes the [[arctan|inverse tangent]] and $\log$ denotes the [[logarithm]]. | where $\mathrm{arctan}$ denotes the [[arctan|inverse tangent]] and $\log$ denotes the [[logarithm]]. | ||
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Revision as of 04:15, 16 May 2016
Theorem: The following formula holds: $$\displaystyle\int \mathrm{arctan}(z) \mathrm{d}z = z\mathrm{arctan}(z) - \dfrac{1}{2}\log(1+z^2)+C,$$ where $\mathrm{arctan}$ denotes the inverse tangent and $\log$ denotes the logarithm.
Proof: █
