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Question Number 138335 by Ñï= last updated on 12/Apr/21

∫(dx/(x^8 +x^4 +1))=?

$$\int\frac{{dx}}{{x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}}=? \\ $$

Answered by MJS_new last updated on 12/Apr/21

x^8 +x^4 +1=  =(x^2 −x+1)(x^2 +x+1)(x^2 −(√3)x+1)(x^2 +(√3)x+1)  now simply decompose  I get  ((√3)/(12))(ln ((x^2 +(√3)x+1)/(x^2 −(√3)x+1)) +2(arctan ((2x−1)/( (√3))) +arctan ((2x+1)/( (√3)))))+C

$${x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}= \\ $$$$=\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}\right) \\ $$$$\mathrm{now}\:\mathrm{simply}\:\mathrm{decompose} \\ $$$$\mathrm{I}\:\mathrm{get} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}\left(\mathrm{ln}\:\frac{{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}}\:+\mathrm{2}\left(\mathrm{arctan}\:\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\mathrm{arctan}\:\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right)+{C} \\ $$

Commented by Ñï= last updated on 12/Apr/21

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

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