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Question Number 134052 by shaker last updated on 27/Feb/21

Answered by mathmax by abdo last updated on 27/Feb/21

$$\mathrm{I}=\int \frac{{\mathrm{x}}^3}{{\mathrm{x}}^6+3}\mathrm{dx}\Rightarrow \mathrm{I}=\int \frac{{\mathrm{x}}^3}{{\mathrm{x}}^6+{\left(3^{\frac{1}{6}}\right)}^6}\mathrm{dx}{=}_{\mathrm{x}=3^{\frac{1}{6}}\mathrm{t}}\int \frac{3^{\frac{1}{2}}{\mathrm{t}}^3}{3(1+{\mathrm{t}}^6)}{3}^{\frac{1}{6}}\mathrm{dt}$$$$=3^{\frac{1}{2}+\frac{1}{6}-1}\int \frac{{\mathrm{t}}^3}{{\mathrm{t}}^6+1}\mathrm{dt}=3^{-\frac{1}{3}}\int \frac{{\mathrm{t}}^3}{{\mathrm{t}}^6+1}\mathrm{dt}$$$${\mathrm{z}}^6+1=0\Rightarrow {\mathrm{z}}^6={\mathrm{e}}^{\mathrm{i}(\pi +2\mathrm{k}\pi )}={\mathrm{e}}^{\mathrm{i}(2\mathrm{k}+1)\pi }\Rightarrow {\mathrm{z}}_{\mathrm{k}}={\mathrm{e}}^{\frac{\mathrm{i}(2\mathrm{k}+1)\pi }{6}}\mathrm{and}$$$$\mathrm{k}\in \left[[0,5]\right]\Rightarrow \frac{{\mathrm{t}}^3}{{\mathrm{t}}^6+1}=\frac{{\mathrm{t}}^3}{\prod _{\mathrm{k}=0}^5(\mathrm{t}-{\mathrm{z}}_{\mathrm{k}})}=\sum _{\mathrm{k}=0}^5\frac{{\mathrm{a}}_{\mathrm{k}}}{\mathrm{t}-{\mathrm{z}}_{\mathrm{k}}}$$$${\mathrm{a}}_{\mathrm{k}}=\frac{{\mathrm{z}}_{\mathrm{k}}^3}{{6\mathrm{z}}_{\mathrm{k}}^5}=\frac{{\mathrm{z}}_{\mathrm{k}}^4}{-6}=-\frac{1}{6}{\mathrm{z}}_{\mathrm{k}}^4\Rightarrow \frac{{\mathrm{t}}^3}{{\mathrm{t}}^6+1}=-\frac{1}{6}\sum _{\mathrm{k}=0}^5\frac{{\mathrm{e}}^{\frac{\mathrm{i}(2\mathrm{k}+1)2\pi }{3}}}{\mathrm{t}-{\mathrm{e}}^{\frac{\mathrm{i}(2\mathrm{k}+1)\pi }{6}}}\Rightarrow$$$$\int \frac{{\mathrm{t}}^3}{1+{\mathrm{t}}^6}\mathrm{dt}=-\frac{1}{6}\sum _{\mathrm{k}=0}^5{\mathrm{e}}^{\frac{2\pi \mathrm{i}(2\mathrm{k}+1)}{3}}\ln (\mathrm{t}-{\mathrm{e}}^{\frac{\mathrm{i}(2\mathrm{k}+1)\pi }{6}})+\mathrm{C}$$

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