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Question Number 147879 by 0731619 last updated on 24/Jul/21

Answered by puissant last updated on 24/Jul/21

$$\mathrm{I}=\int \frac{1}{{\mathrm{x}}^5+1}\mathrm{dx}$$$$=\int \frac{\mathrm{dx}}{(\mathrm{x}+1)({\mathrm{x}}^4-{\mathrm{x}}^3+{\mathrm{x}}^2-\mathrm{x}+1)}$$$$=\int \frac{\frac{1}{5}}{\mathrm{x}+1}\mathrm{dx}+\int \frac{-\frac{1}{5}{\mathrm{x}}^3+\frac{2}{5}{\mathrm{x}}^2-\frac{3}{5}\mathrm{x}+\frac{4}{5}}{{\mathrm{x}}^4-{\mathrm{x}}^3+{\mathrm{x}}^2-\mathrm{x}+1}\mathrm{dx}$$$$=\frac{1}{5}\ln \mid \mathrm{x}+1\mid -\frac{1}{5}\int \frac{{\mathrm{x}}^3-{2\mathrm{x}}^2+3\mathrm{x}-4}{{\mathrm{x}}^4-{\mathrm{x}}^3+{\mathrm{x}}^2-\mathrm{x}+1}\mathrm{dx}$$$$=\frac{1}{5}\ln \mid \mathrm{x}+1\mid -\frac{1}{20}\int \frac{{4\mathrm{x}}^3-{8\mathrm{x}}^2+12\mathrm{x}-16}{{\mathrm{x}}^4-{\mathrm{x}}^3+{\mathrm{x}}^2-\mathrm{x}+1}\mathrm{dx}$$$$=\frac{1}{5}\ln \mid \mathrm{x}+1\mid -\frac{1}{20}\ln \mid {\mathrm{x}}^4-{\mathrm{x}}^3+{\mathrm{x}}^2-\mathrm{x}+1\mid +\frac{1}{20}\int \frac{{5\mathrm{x}}^2-14\mathrm{x}+15}{{\mathrm{x}}^4-{\mathrm{x}}^3+{\mathrm{x}}^2-\mathrm{x}+1}\mathrm{dx}$$$$\mathrm{let}\mathrm{Q}=\frac{1}{20}\int \frac{{5\mathrm{x}}^2-14\mathrm{x}+15}{{\mathrm{x}}^4-{\mathrm{x}}^3+{\mathrm{x}}^2-\mathrm{x}+1}\mathrm{dx}$$$$=\frac{1}{20}\int \frac{{5\mathrm{x}}^2-14\mathrm{x}+15}{({\mathrm{x}}^2+\left(\frac{\sqrt{5}-1}{2}\right)\mathrm{x}+1)({\mathrm{x}}^2-\left(\frac{\sqrt{5}+1}{2}\right)\mathrm{x}+1)}\mathrm{dx}$$$$=\frac{1}{20}\int \frac{\mathrm{Ax}+\mathrm{B}}{({\mathrm{x}}^2+\left(\frac{\sqrt{5}-1}{2}\right)\mathrm{x}+1)}\mathrm{dx}+\frac{1}{20}\int \frac{\mathrm{Cx}+\mathrm{D}}{({\mathrm{x}}^2-\left(\frac{\sqrt{5}+1}{2}\right)\mathrm{x}+1}\mathrm{dx}$$$$\mathrm{J}=\frac{1}{20}\int \frac{\mathrm{Ax}+\mathrm{b}}{{\mathrm{x}}^2+\left(\frac{\sqrt{5}-1}{2}\right)\mathrm{x}+1}\mathrm{dx}$$$$=\frac{1}{20}\int \frac{2\sqrt{5}\mathrm{x}+\frac{13+15\sqrt{5}}{2\sqrt{5}}}{{\mathrm{x}}^2+\left(\frac{\sqrt{5}-1}{2}\right)\mathrm{x}+1}\mathrm{dx}=\frac{1}{4\sqrt{5}}\int \frac{2\mathrm{x}+\frac{13+15\sqrt{5}}{10}}{{\mathrm{x}}^2+\left(\frac{\sqrt{5}-1}{2}\right)\mathrm{x}+1}\mathrm{dx}$$$$=\frac{1}{4\sqrt{5}}\int \frac{2\mathrm{x}+\frac{\sqrt{5}-1}{2}}{{\mathrm{x}}^2+\left(\frac{\sqrt{5}-1}{2}\right)\mathrm{x}+1}\mathrm{dx}+\frac{1}{4\sqrt{5}}\int \frac{10\sqrt{5}+18}{{\mathrm{x}}^2+\left(\frac{\sqrt{5}-1}{2}\right)\mathrm{x}+1}\mathrm{dx}$$$$=\frac{1}{4\sqrt{5}}\ln \mid {\mathrm{x}}^2+\left(\frac{\sqrt{5}-1}{2}\right)\mathrm{x}+1\mid +\frac{10\sqrt{5}+18}{40\sqrt{5}}\int \frac{\mathrm{dx}}{{(\mathrm{x}+\frac{\sqrt{5}-1}{2})}^2+{\left(\sqrt{\frac{10+2\sqrt{5}}{16}}\right)}^2}$$$$=\frac{1}{4\sqrt{5}}\ln \mid {\mathrm{x}}^2+\left(\frac{\sqrt{5}-1}{2}\right)\mathrm{x}+1\mid +\frac{10\sqrt{5}+18}{40\sqrt{5}}\times \frac{4}{\sqrt{10+2\sqrt{5}}}\arctan \left(\frac{4\mathrm{x}+\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)+\mathrm{C}$$$$\mathrm{De}\mathrm{facon}\mathrm{analogue},\mathrm{on}\mathrm{trouve}$$$$\frac{1}{20}\int \frac{\mathrm{Cx}+\mathrm{D}}{{\mathrm{x}}^2-\left(\frac{\sqrt{5}+1}{2}\right)\mathrm{x}+1}\mathrm{dx}$$$$=-\frac{1}{4\sqrt{5}}\ln \mid {\mathrm{x}}^2-\left(\frac{\sqrt{5}-1}{2}\right)\mathrm{x}+1\mid -\frac{10\sqrt{5}+18}{40\sqrt{5}}\times \frac{4}{\sqrt{10-2\sqrt{5}}}\arctan \left(\frac{4\mathrm{x}-\sqrt{5}-1}{\sqrt{10-2\sqrt{5}}}\right)+\mathrm{C}$$$$\mathrm{apres}\mathrm{sommation},\mathrm{on}\mathrm{trouve}:$$$$$$$$\mathrm{I}=\frac{1}{5}\ln \mid \mathrm{x}+1\mid -\frac{1}{20}\ln \mid {\mathrm{x}}^4-{\mathrm{x}}^3+{\mathrm{x}}^2-\mathrm{x}+1\mid +\frac{1}{4\sqrt{5}}\ln \mid {\mathrm{x}}^2+\left(\frac{\sqrt{5}-1}{2}\right)\mathrm{x}+1\mid$$$$+\frac{10\sqrt{5}+18}{10\sqrt{5}}\times \frac{1}{\sqrt{10+2\sqrt{5}}}\arctan \left(\frac{4\mathrm{x}+\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)-\frac{1}{4\sqrt{5}}\ln \mid {\mathrm{x}}^2-\left(\frac{\sqrt{5}+1}{2}\right)\mathrm{x}+1\mid$$$$+\frac{10\sqrt{5}-18}{10\sqrt{5}}\times \frac{1}{\sqrt{10-2\sqrt{5}}}\arctan \left(\frac{4\mathrm{x}-\sqrt{5}-1}{2}\right)+\mathrm{K}...$$

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