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find-dx-1-x-6-




Question Number 70686 by mathmax by abdo last updated on 06/Oct/19
find ∫ (dx/(1+x^6 ))
$${find}\:\int\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{6}} } \\ $$
Answered by kaivan.ahmadi last updated on 06/Oct/19
Answered by MJS last updated on 07/Oct/19
∫(dx/(x^6 +1))=  =(1/3)∫(dx/(x^2 +1))+(1/6)∫(((√3)x+2)/(x^2 +(√3)x+1))dx−(1/6)∫(((√3)x−2)/(x^2 −(√3)x+1))dx=  =(1/3)arctan x +       +((√3)/(12))ln (x^2 +(√3)x+1) +(1/6)arctan (2x+(√3)) −       −((√3)/(12))ln (x^2 −(√3)x+1) +(1/6)arctan (2x−(√3)) +C
$$\int\frac{{dx}}{{x}^{\mathrm{6}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\sqrt{\mathrm{3}}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}}{dx}−\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\sqrt{\mathrm{3}}{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arctan}\:{x}\:+ \\ $$$$\:\:\:\:\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{arctan}\:\left(\mathrm{2}{x}+\sqrt{\mathrm{3}}\right)\:− \\ $$$$\:\:\:\:\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{arctan}\:\left(\mathrm{2}{x}−\sqrt{\mathrm{3}}\right)\:+{C} \\ $$

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