<s>[INST] Partial fractions integral. Compute $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$ 
Evaluate $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$.

I directly approached this with partial fractions and rewritten $x^6+1=(x^2)^3+1=(x^2+1)(x^4+x^2+1)$.
Therefore the integral is:
$$\int_{0}^{1} \frac {x^4+1}{(x^2+1)(x^4+x^2+1)}dx=-2\int_{0}^{1} \frac {x}{1+x^2}dx+2\int_{0}^{1}\frac 1{1+x^2}dx-\int_{0}^1\frac {(x-1)^2}{(x^2+1)^2+(\frac {\sqrt{3}}2)^2}dx$$
But I don't know how to solve the last term.
I answered the question... I realized my silly mistake, sorry for bothering...
 [/INST] $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
\int_{0}^{1}{x^{4} + 1 \over x^{6} + 1}\,\dd x & =
\int_{0}^{1}{x^{4} + 1 - x^{10} - x^{6} \over 1 - x^{12}}\,\dd x =
{1 \over 12}\int_{0}^{1}{x^{-7/12} + x^{-11/12} - x^{-1/12} - 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^{-7/12} \over 1 - x}\,\dd x}
\\[2mm] & +
{1 \over 12}\pars{\int_{0}^{1}{1 - x^{-1/12} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{-11/12} \over 1 - x}\,\dd x}
\\[5mm] & =
{\pars{H_{-5/12} - H_{-7/12}} +
\pars{H_{-1/12} - H_{-11/12}}\over 12}\qquad\pars{~H_{z}:\ Harmonic\ Number~}
\\[5mm] & =
{\pi\cot\pars{5\pi/12} + \pi\cot\pars{\pi/12}\over 12}
\qquad\pars{~Euler\ Reflection\ Formula~}
\\[5mm] & =
{\pi \over 12}\,{\sin\pars{\pi/2} \over \sin\pars{5\pi/12}\sin\pars{\pi/12}} =
{\pi \over 12}\,{1 \over \bracks{\cos\pars{\pi/3} - \cos\pars{\pi/2}}/2} =
{\pi \over 6}\,{1 \over 1/2 - 0}
\\[5mm] & = \bbx{\pi \over 3} \approx 1.0472
\end{align}
</s>