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...

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 $x-\dfrac1x=u$
