How do I integrate $\frac{1}{x^6+1}$ My technique so far was substitution with the intent of getting to a sum of three fractions with squares in their denominators.
$t = x^2 \\
\frac{1}{x^6 + 1} = \frac{1}{t^3+1} = \frac{1}{(t+1)(t^2-t+1)}$
Then I try to reduce this fraction into a sum of two fractions
$\frac{A}{t+1} + \frac{B}{t^2-t+1} = \frac{(At^2-(A-B)t) + A + B}{(t+1)(t^2-t+1)}$
And this is where I reach a dead-end
$\begin{cases}
   At^2-(A-B)t &= 0 \\
   A + B &= 1
  \end{cases}$
Any techniques I'm overlooking?

You have to find the (complex) pairs of roots of the polynomial $x^6+1$, and then use the uncertain coefficients method as you did.
Notice that using the substitution $t=x^2$ leaves you with $x=\sqrt{t}$ and $dx=dt/\sqrt{t}$ which doesn't look quite nicely integrable.
When you have the polynomials $t^2+pt+q$ with complex roots, you can use a clever linear substitution to change that into $u^2+1$; then $\int du/(u^2+1) = \arctan u+C$.
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As for finding the roots of $x^6+1=0$, we can either guess them or help ourselves. You already know that $x^2=t$ and $t^3+1=0$, which could be helpful. But we may even notice that $(x^6+1)(x^6-1)=x^{12}-1$, therefore $x$ are the 12th roots of unity which are not 6th roots of unity. These are
$$\xi, \xi^3, \xi^5, \xi^7=\overline{\xi^5}, \xi^9=\overline{\xi^3}\text{ and }\xi^{11}=\overline{\xi},$$
where $\xi=e^{2\pi i/12}=e^{i\pi/6}$ is the fundamental (or first) 12th root of unity.
