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...
$$\begin{align} \int_0^1\frac{1}{1+x^6} \,\mathrm dx &=\frac{1}{2}\left[ \int_0^1 \frac{(1-x^2+x^4)+x^2+(1-x^4)}{(1+x^2)(1-x^2+x^4)} \,\mathrm dx \right]\tag{1}\\ &=\frac{1}{2}\left[\int_0^1 \frac{1}{1+x^2} \,\mathrm dx+ \int_0^1 \frac{x^2}{1+x^6} \,\mathrm dx + \color{grey}{\int_0^1 \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx}\right] \tag{2}\\ &=\frac{1}{2}\left[\frac\pi4+ \frac\pi{12} +\frac{\log(2+\sqrt{3})}{\sqrt{3}} \right] \tag{3}\\ &=\frac12\left[\frac{\pi+\sqrt3\log(2+\sqrt{3})}{3} \right] \tag{4}\\ \end{align}$$
$$\int_0^1\frac{1}{1+x^6} \,\mathrm dx =\frac{\pi+\sqrt3\log(2+\sqrt{3})}{6}$$
$\text{Explanation : }(3)$ Substituting $\displaystyle t=x+\frac1x\iff \,\mathrm dt=\left(1-\frac1{x^2}\right)\,\mathrm dx$ in last integral
$$\begin{align} \color{grey}{J} &=\color{grey}{\int_0^1 \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx} =\int_0^1 \frac{\frac{1}{x^2}-1}{x^2-1+\frac{1}{x^2}}\,\mathrm dx =\int_2^\infty \frac{1}{t^2-3}\,\mathrm dt\\ &=\frac{1}{2\sqrt{3}}\log\left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right) =\color{grey}{\frac{\log(2+\sqrt{3})}{\sqrt{3}}}\\ \end{align}$$