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How do you prove that

$$\int_{0}^{1}\dfrac{dx}{1+x^6}=\frac{\pi+\sqrt3\log(2+\sqrt3)}{6}$$

My steps: First sub $\displaystyle u=x^3, \sqrt[3]u=x, dx=\dfrac{u^{-2/3}}{3} du\implies\dfrac{1}{3}\int_{0}^{1}\dfrac{u^{-2/3}}{1+u^2}$

Then sub $\displaystyle u=\tan\theta, du=\sec^2\theta d\theta\implies \dfrac{1}{3}\int_{0}^{\pi/4}\sec^{-4/3}\theta d\theta$

After which I am stuck...

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  • $\begingroup$ Partial Fractions should work well enough here. $1+x^6 = 1+(x^2)^3 = (1+x^2)(1-x^2+x^4)$, etc... $\endgroup$
    – Steven Stadnicki
    Dec 9, 2014 at 2:16
  • $\begingroup$ Ends up with a simple atan substitution and $\int\dfrac{2-x^2}{3(x^4-x^2+1)}$ $\endgroup$
    – Teoc
    Dec 9, 2014 at 2:21

1 Answer 1

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$$\begin{align} \int_0^1\frac{1}{1+x^6} \,\mathrm dx &=\frac{1}{2}\left[ \int_0^1 \frac{(1-x^2+x^4)+x^2+(1-x^4)}{(1+x^2)(1-x^2+x^4)} \,\mathrm dx \right]\tag{1}\\ &=\frac{1}{2}\left[\int_0^1 \frac{1}{1+x^2} \,\mathrm dx+ \int_0^1 \frac{x^2}{1+x^6} \,\mathrm dx + \color{grey}{\int_0^1 \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx}\right] \tag{2}\\ &=\frac{1}{2}\left[\frac\pi4+ \frac\pi{12} +\frac{\log(2+\sqrt{3})}{\sqrt{3}} \right] \tag{3}\\ &=\frac12\left[\frac{\pi+\sqrt3\log(2+\sqrt{3})}{3} \right] \tag{4}\\ \end{align}$$

$$\int_0^1\frac{1}{1+x^6} \,\mathrm dx =\frac{\pi+\sqrt3\log(2+\sqrt{3})}{6}$$

$\text{Explanation : }(3)$ Substituting $\displaystyle t=x+\frac1x\iff \,\mathrm dt=\left(1-\frac1{x^2}\right)\,\mathrm dx$ in last integral

$$\begin{align} \color{grey}{J} &=\color{grey}{\int_0^1 \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx} =\int_0^1 \frac{\frac{1}{x^2}-1}{x^2-1+\frac{1}{x^2}}\,\mathrm dx =\int_2^\infty \frac{1}{t^2-3}\,\mathrm dt\\ &=\frac{1}{2\sqrt{3}}\log\left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right) =\color{grey}{\frac{\log(2+\sqrt{3})}{\sqrt{3}}}\\ \end{align}$$

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    $\begingroup$ Very nice & elementary. (+1) $\endgroup$
    – Venus
    Dec 9, 2014 at 3:37
  • $\begingroup$ Nice trick? What reasoning did you use to choose such a method? Or is this a common trick that I'm naive to? $\endgroup$
    – mathamphetamines
    Dec 9, 2014 at 5:22
  • $\begingroup$ It's well known here see: math.stackexchange.com/questions/985837/… $\endgroup$
    – FDP
    Dec 9, 2014 at 13:26
  • $\begingroup$ @FDP Actually I've deleted my comment answering matha..'s question! $\endgroup$
    – Aditya Hase
    Dec 9, 2014 at 13:44
  • $\begingroup$ Nice ! Someone have added a link to the question about the computation of $\displaystyle \int_{-\infty}^{+\infty}\dfrac{1}{1+x^4}dx$ $\endgroup$
    – FDP
    Dec 9, 2014 at 13:57

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