Evaluate $\int \frac 1{x^{12}+1} \, dx$ 
Evaluate $\displaystyle \int \frac 1{x^{12}+1} \, dx$

I tried writing this in partial fractions.
$$\int \frac 1{x^{12}+1} \, dx=\int \frac{1}{[(x^6+1)+\sqrt{2}x^3][(x^6+1)-\sqrt{2}x^3]} \, dx$$
So I did:
\begin{align}
\frac{1}{[(x^6+1)+\sqrt{2}x^3][(x^6+1)-\sqrt{2}x^3]} &= \frac{Ax^5+Bx^4+Cx^3+Dx^2+Ex+F}{[(x^6+1)-\sqrt{2}x^3]}\\&{}\qquad+\frac{A_1x^5+B_1x^4+C_1x^3+D_1x^2+E_1x+F_1}{[(x^6+1)+\sqrt{2}x^3]}\end{align}
After messy work, I found $A=A_1=B=B_1=D=D_1=E=E_1=0$ and
$$C=-\frac 1{2\sqrt{2}},\qquad C_1=\frac 1{2\sqrt{2}},\qquad F=F_1=\frac 12$$
I now get the integral
$$\int  \frac{-\frac 1{2\sqrt{2}}x^3+\frac 12}{(x^6+1)-\sqrt{2}x^3}\, dx + \int \frac{\frac 1{2\sqrt{2}}x^3+\frac 12}{(x^6+1)+\sqrt{2}x^3} \, dx$$
But how can I go from here? I am stuck.

The denominator of the fraction has twelve complex roots which are of the form $\omega^k$, where $\omega$ is a twelfth root of minus one.
By decomposition in simple fractions,
$$\frac1{x^{12}+1}=\frac1{12}\left(\frac{\omega^1}{x-\omega^1}+\frac{\omega^3}{x-\omega^3}+\cdots\frac{\omega^{23}}{x-\omega^{23}}\right).$$
Now, as
$$\frac1{x-\omega}=\frac1{x-r-is}=\frac{x-r+is}{x^2-2rx+r^2+s^2}=\frac{y+is}{y^2+s^2}.$$
we can integrate to obtain
$$\log(x-\omega)=\log\sqrt{y^2+s^2}+i\arctan\frac ys=\log\sqrt{(x-r)^2+s^2}+i\arctan\frac {x-r}s$$
and you can sum the twelve contributions. By symmetry, they
can be grouped in pairs and you get a sum of six logarithms and six arc-tangents.
