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Motivated by the final step of my answer in evaluating this integral, I tried to make up an indefinite integral on these lines:

$$\int\frac{x^6-x^5-x+1}{x^8+x^4+1}\mathrm dx$$

My solution:

Since the denominator of the integrand can be factorized as $$(x^4-x^2+1)(x^2-x+1)(x^2+x+1)=(x^6-x^5+x^3-x+1)(x^2+x+1)$$

The integral can be split into two:

$$\begin{align}I&=\int\frac{\mathrm dx}{x^2+x+1}-\frac14\int\frac{\mathrm d(x^4)}{x^8+x^4+1}\\&=\frac2{\sqrt3}\tan^{-1}\left(\frac{2x+1}{\sqrt3}\right)-\frac1{2\sqrt3}\tan^{-1}\left(\frac{2x^4+1}{\sqrt3}\right)+C\end{align}$$

Is there any other method to evaluate this intriguing integral? Any comments or ideas are highly appreciated.

Integreek
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    The substitution $x = e^{iu}$ gives $I = 2i\int\frac{\cos(3u)-\cos(2u)}{2\cos(4u)+1} ~du$, but I haven't figured out where to go from there. – Dan Apr 17 '25 at 15:49

2 Answers2

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$$\frac{x^6-x^5-x+1}{x^8+x^4+1}=\frac{1}{4 \left(x^2-x+1\right)}+\frac{3}{4 \left(x^2+x+1\right)}-\frac{x}{2 \left(x^4-x^2+1\right)}$$ The first and second terms do not make any problem. For the third $$\frac{x}{x^4-x^2+1}=\frac x{(x^2-a)(x^2-b)}$$ $$\frac{x}{x^4-x^2+1}=\frac 1{a-b}\Big(\frac x{x^2-a}-\frac x{x^2-b} \Big)$$ $$\int\frac{x}{x^4-x^2+1}\,dx=\frac 1{2(a-b)}\log \left(\frac{x^2-a}{x^2-b}\right) $$

where $$a=\frac{1+i \sqrt{3}}{2}\qquad \text{and}\qquad b=\frac{1-i \sqrt{3}}{2}$$

Claude Leibovici
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As suggested by @Dan,

$$\begin{align}\int\frac{x^6-x^5-x+1}{x^8+x^4+1}\mathrm dx&=2i\int\frac{(\cos u-\frac14)(4\cos^2u-3)-\frac14(4\cos^2u-1)}{(4\cos^2u-3)(4\cos^2u-1)}\mathrm du\\&=2i\int\frac{\cos u~\mathrm du}{3-4\sin^2u}-\frac{i}4\int\frac{\mathrm d(\tan u)}{1-3\tan^2u}-\frac{i}4\int\frac{\mathrm d(\tan u)}{3-\tan^2u}\\&=\frac{i}{2\sqrt3}\log\left(\frac{x^2+\sqrt3ix-1}{x^2-\sqrt3ix-1}\right)+\frac{i}{8\sqrt3}\log\left(\frac{4ix^4-4\sqrt3x^2-4i}{4ix^4+4\sqrt3x^2-4i}\right)+C\end{align}$$

Integreek
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