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 $\pi/4$
 A: You can take the long way using partial fractions, which gives you:
$$\underbrace{\frac{1}{3(x^2+1)}}_{f_1(x)}+\underbrace{\frac{\sqrt 3 x+2}{6(x^2+\sqrt3 x+1)}}_{f_2(x)}+\underbrace{\frac{3\sqrt3x-6}{18(-x^2+\sqrt3x-1)}}_{f_3(x)}$$
$$\int_0^1f_1(x)dx=\frac 13\int_0^1\frac 1{1+x^2}dx=\left[\arctan(x)\right]_0^1$$
$$\int_0^1f_2(x)dx=\int_0^1\frac{\sqrt 3 x+2}{6(x^2+\sqrt3 x+1)}dx=\frac 16\int_0^1\frac{\sqrt3x+1}{x^2+\sqrt3 x+1}+\frac{1}{x^2+\sqrt3 x+1}$$
From here on out it's pretty straightforward. The answer will involve logarithms and $\arctan$'s. It's similar for $f_3(x)$...
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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The exponent $\ds{2 \over 3}$ in the $\ds{\cot}$-function makes it rather cumbersome. You can go to other straightforward avenues. 

$\ds{\color{#f00}{\int_{0}^{1}{\dd x \over x^{6} + 1}}:\ ?}$.

1.\begin{align}
\color{#f00}{\int_{0}^{1}{\dd x \over x^{6} + 1}} & =
\int_{0}^{1}{1 - x^{6} \over 1 - x^{12}}\,\dd x =
{1 \over 12}\int_{0}^{1}{x^{-11/12}\,\,\, -\,\,\, x^{-5/12} \over 1 - x}\,\dd x \\[5mm] & =
{1 \over 12}\pars{\int_{0}^{1}{1 - x^{-5/12} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{-11/12} \over 1 - x}\,\dd x}
\\[5mm] & =
{1 \over 12}\bracks{\Psi\pars{7 \over 12} - \Psi\pars{1 \over 12}}
\qquad\pars{~\Psi:\ Digamma\ Function~}
\end{align}
With $\ds{\pars{~p = 1,2,3\ldots\,,\ q = 2,3,4\ldots\,,\ p < q~}}$,
$\ds{\Psi\pars{p \over q}}$ is evaluated with (  see G & R,
$\ds{\mathbf{8.363}}$.6 ) the identity:
$$
\Psi\pars{p \over q} =
-\gamma -\ln\pars{2q} - \half\,\pi\cot\pars{\pi p \over q} +
2\sum_{k = 1}^{\left\lfloor\,{\pars{q + 1}/2}\,\right\rfloor - 1}
\cos\pars{2\pi kp \over q}\ln\pars{\sin\pars{\pi k \over q}}
$$
$\ds{\gamma}$ is the Euler-Mascheroni Constant.

The final result becomes:
$$
\color{#f00}{\int_{0}^{1}{\dd x \over x^{6} + 1}} =
\color{#f00}{{1 \over 6}\bracks{\pi + \root{3}\ln\pars{2 + \root{3}}}} \approx 0.9038
$$
A: It is easier to solve this integral using partiel fraction decomposition
$\frac{1}{1+x^6} =\frac{1}{3(1+x^2)}+\frac{2-x^2}{3(x^4-x^2+1)}$
and solving the 3 integrals seperatly.
A: Hint:
note that 
$$
x^6+1=(x^2+1)(x^2-\sqrt{3}+1)(x^2+\sqrt{3}+1)
$$
than use partial fraction decomposition.
A: 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.
A: Continuing where MK12 left off, we proceed as follows:
$$\frac{2-x^2}{x^4-x^2+1} = \frac{1}{2} \times \frac{4-2x^2}{x^4-x^2+1} = \frac{1}{2} \times \frac{1+x^2 + 3(1-x^2)}{x^4-x^2+1}$$
Split the sub-integral into two parts. For one, make the substitution $u=x+\frac{1}{x}$ and the other $v = x-\frac{1}{x}$
Then the rest should be fairly trivial, provided you are meticulously careful with the limits of integration.
