<s>[INST] 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.
 [/INST] $x^{12}+1=(x^4+1)(x^8-x^4+1)$ and $x^4+1=(x^2+x\sqrt{2}+1)(x^2-\sqrt{2}x+1)$
further we have $x^8-x^4+1=- \left( \sqrt {2}{x}^{3}+{x}^{4}+x\sqrt {2}+{x}^{2}+1 \right) 
 \left( \sqrt {2}{x}^{3}-{x}^{4}+x\sqrt {2}-{x}^{2}-1 \right) 
$
</s>