Difference between revisions of "Antiderivative of arctan"
From specialfunctionswiki
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$\displaystyle\int \mathrm{arctan}(z) \mathrm{d}z = z\mathrm{arctan}(z) - \dfrac{1}{2}\log \left(1+z^2 \right)+C,$$ | $$\displaystyle\int \mathrm{arctan}(z) \mathrm{d}z = z\mathrm{arctan}(z) - \dfrac{1}{2}\log \left(1+z^2 \right)+C,$$ | ||
where $\mathrm{arctan}$ denotes the [[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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| − | + | ==Proof== | |
| − | + | ||
| − | + | ==References== | |
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.
