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...
 A: Set $x^2=y$ $$\dfrac{y^2+1}{y^3+1}=\dfrac A{y+1}+\dfrac{By+C}{y^2-y+1}$$
$$y^2=y^2(A+B)+y(B+C-A)+A+C$$
$\implies A+C=1\iff C=1-A, B+C=A\implies B=A-C=2A-1,$
$1=A+B=3A-1\iff A=\dfrac23, B=\dfrac13,C=\dfrac13$
$$\implies\dfrac{x^4+1}{x^6+1}=\dfrac1{3(x^2+1)}+\dfrac{x^2+1}{3(x^4-x^2+1)}$$
Now $\dfrac{x^2+1}{x^4-x^2+1}=\dfrac{1+\dfrac1{x^2}}{\left(x-\dfrac1x\right)^2+1}$
set $x-\dfrac1x=u$
A: Note that $x^6+1=(x^2)^3+1=(x^2+1)(x^4-x^2+1)$ and
use the fact that you are integrating over $[0,1]$,
$$\int_0^1 \frac{x^4+1\pm x^2}{x^6+1}dx =\int_0^1 \frac{x^4-x^2+1}{x^6+1}dx+ \frac{1}{3}\int_0^1 \frac{3x^2}{x^6+1}dx\\=\int_0^1 \frac{1}{x^2+1}dx+ \frac{1}{3}\int_0^1 \frac{1}{t^2+1}dt=\left(1+\frac{1}{3}\right)\frac{\pi}{4}=\frac{\pi}{3}$$
where, in the second integral, $t=x^3$.
A: I'm sorry but I had a really stupid mistake and now I solved the integral really easily...
So this is what I did:
$$I=\int_0^1 \frac {x^4+1}{(x^2)^3+1}dx=\int_0^1\frac{(x^2+1)^2-2x^2}{(x^2+1)(x^4-x^2+1)}dx$$
then we divide top and bottom by $x^2$ and separate the integrals...
we get:
$$I=\int_0^1\frac{1+\frac 1{x^2}}{(x-\frac 1x)^2+1}dx-2\int_0^1\frac{x^2}{(x^3)^2+1}dx$$
and this is:
$$I=arctan(x-\frac 1x)-\frac 23arctan(x^3)$$
which from $0$ to $1$, we get: $I=\frac \pi3.$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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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}
