Difference between revisions of "Antiderivative of arctan"
From specialfunctionswiki
| Line 1: | Line 1: | ||
<div class="toccolours mw-collapsible mw-collapsed"> | <div class="toccolours mw-collapsible mw-collapsed"> | ||
<strong>[[Antiderivative of arctan|Theorem]]:</strong> The following formula holds: | <strong>[[Antiderivative of arctan|Theorem]]:</strong> The following formula holds: | ||
| − | $$\displaystyle\int \mathrm{arctan}(z) \mathrm{d}z = z\mathrm{arctan}(z) - \dfrac{1}{2}\log(1+z^2)+C,$$ | + | $$\displaystyle\int \mathrm{arctan}(z) \mathrm{d}z = z\mathrm{arctan}(z) - \dfrac{1}{2}\log \left(1+z^2 \right)+C,$$ |
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]]. | ||
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
Revision as of 04:16, 16 May 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: █
