<s>[INST] Solve $\int_{0}^{1}\frac{1}{1+x^6} dx$ Let $$x^3 = \tan y\ \ \text{ so that }\ x^2 = \tan^{2/3}y$$
$$3x^2dx = \sec^2(y)dy$$
$$\int_{0}^{1}\frac{1}{1+x^6}dx = \int_{1}^{\pi/4}\frac{1}{1+\tan^2y}\cdot \frac{\sec^2y}{3\tan^{2/3}y}dy = \frac{1}{3}\int_{1}^{\pi/4} \cot^{2/3}y\ dy$$
How should I proceed after this?
EDITED: Corrected the final integral and the limit from $45$ to $\pi/4$
 [/INST] By writing the integrand function as its Taylor series centered at $x=0$ and performing termwise integration we get:
$$ I=\int_{0}^{1}\frac{dx}{1+x^6}=\sum_{k\geq 0}\frac{(-1)^k}{6k+1}=\sum_{n\geq 0}\left(\frac{1}{12k+1}-\frac{1}{12k+7}\right).\tag{1} $$
The last series can now be computed through the discrete Fourier transform (DFT).
Let $\left\{\omega=\exp\left(\frac{\pi i}{6}\right),\omega^5,\omega^7,\omega^{11}\right\}$ be the set of the primitive twelth roots of unity and $\chi(n)$ be the arithmetic function that equals $+1$ if $n\equiv\!1\!\!\pmod{12}$, $-1$ if $n\equiv\!7\!\!\pmod{12}$ and zero otherwise.
We have:
$$\chi(n) = \frac{1}{12}\left(-2\omega^5 \omega^n -2i \omega^{3n}-2\omega \omega^{5n}+2\omega^5\omega^{7n}+2i\omega^{9n}+2\omega \omega^{11n}\right)\tag{2}$$
and for every $k\in\{1,2,\ldots,11\}$ the equality:
$$ \sum_{n\geq 1}\frac{\omega^{kn}}{n}=-\log(1-\omega^k)\tag{3} $$
holds. By putting everything together,
$$ I = \sum_{n\geq 0}\left(\frac{1}{12k+1}-\frac{1}{12k+7}\right) =\sum_{n\geq 1}\frac{\chi(n)}{n}= \color{red}{\frac{\pi+\sqrt{3}\log(2+\sqrt{3})}{6}}\tag{4} $$
easily follows from $2+\sqrt{3}=\cot\frac{\pi}{12}$.

Another approach is to use the substitution $\frac{1}{1+x^6}=y$ to get:
$$ I = \frac{1}{6}\int_{\frac{1}{2}}^{1}y^{-1/6}(1-y)^{-5/6}\,dy \tag{5}$$
then, by setting 
$$ J=\frac{1}{6}\int_{0}^{\frac{1}{2}}y^{-1/6}(1-y)^{-5/6}\,dy = \frac{1}{6}\int_{\frac{1}{2}}^{1}(1-y)^{-1/6}y^{-5/6}\,dy \tag{6} $$
noticing that:
$$ I+J=\frac{1}{6}\int_{0}^{1}y^{-1/6}(1-y)^{-5/6}\,dy=\frac{\pi}{3}\tag{7}$$
by Euler's beta function and the reflection formulas for the $\Gamma$ function, while $I-J$ provides the logarithmic contribute.
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