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Question Number 205928 by mathzup last updated on 03/Apr/24

$$calculate{\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^4+x^8}$$

Answered by Frix last updated on 03/Apr/24

$$\int \frac{dx}{x^8+x^4+1}=$$$$=\frac{1}{4}\int \frac{dx}{x^2-x+1}+$$$$+\frac{1}{4}\int \frac{dx}{x^2+x+1}-$$$$-\frac{\sqrt{3}}{12}\int \frac{2x-\sqrt{3}}{x^2-\sqrt{3}x+1}dx+$$$$+\frac{\sqrt{3}}{12}\int \frac{2x+\sqrt{3}}{x^2+\sqrt{3}x+1}dx=$$$$=\frac{\sqrt{3}}{6}{\tan }^{-1}\frac{\sqrt{3}(2x-1)}{3}+$$$$+\frac{\sqrt{3}}{6}{\tan }^{-1}\frac{\sqrt{3}(2x+1)}{3}-$$$$-\frac{\sqrt{3}}{12}\ln (x^2-\sqrt{3}x+1)+$$$$+\frac{\sqrt{3}}{12}\ln (x^2+\sqrt{3}x+1)$$$$\Rightarrow$$$$\mathrm{Answer}\mathrm{is}\frac{\sqrt{3}}{6}\pi$$

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