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Question Number 129645 by liberty last updated on 17/Jan/21

$$ϝ=\int \frac{dx}{1+x^{12}}.$$

Answered by mathmax by abdo last updated on 17/Jan/21

$$\mathrm{roots}\mathrm{of}{\mathrm{z}}^{12}+1=0\Rightarrow {\mathrm{z}}^{12}={\mathrm{e}}^{\mathrm{i}\pi +2\mathrm{ik}\pi }={\mathrm{e}}^{\mathrm{i}(2\mathrm{k}+1)\pi }\Rightarrow {\mathrm{z}}_{\mathrm{k}}={\mathrm{e}}^{\mathrm{i}\frac{2\mathrm{k}+1)\pi }{12}}\mathrm{withk}\in \left[[0,11]\right]$$$$\mathrm{and}\frac{1}{{\mathrm{z}}^{12}+1}=\frac{1}{\prod _{\mathrm{k}=0}^{11}(\mathrm{z}-{\mathrm{z}}_{\mathrm{k}})}=\sum _{\mathrm{k}=0}^{11}\frac{{\mathrm{a}}_{\mathrm{k}}}{\mathrm{z}-{\mathrm{z}}_{\mathrm{k}}}$$$${\mathrm{a}}_{\mathrm{k}}=\frac{1}{{12\mathrm{z}}_{\mathrm{k}}^{11}}=\frac{{\mathrm{z}}_{\mathrm{k}}}{-12}\Rightarrow {\mathrm{a}}_{\mathrm{k}}=-\frac{1}{12}\sum _{\mathrm{k}=0}^{11}\frac{{\mathrm{z}}_{\mathrm{k}}}{\mathrm{z}-{\mathrm{z}}_{\mathrm{k}}}\Rightarrow$$$$\mathrm{F}=-\frac{1}{12}\sum _{\mathrm{k}=0}^{11}{\mathrm{z}}_{\mathrm{k}}\int \frac{\mathrm{dx}}{\mathrm{x}-{\mathrm{z}}_{\mathrm{k}}}=-\frac{1}{12}\sum _{\mathrm{k}=0}^{11}{\mathrm{z}}_{\mathrm{k}}\ln (\mathrm{x}-{\mathrm{z}}_{\mathrm{k}})+\mathrm{C}$$

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