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?

Observe that $$x^6 + 1 = (x^2)^3 + 1 = (x^2 + 1)(x^4 - x^2 + 1).$$  To factor the quartic term, suppose there is a factorization of the form $$\begin{align*} x^4 - x^2 + 1 &= (x^2 + ax + 1)(x^2 + bx + 1) \\ &= x^4 + (a+b)x^3 + (ab+2)x^2 + (a+b)x + 1. \end{align*}$$  We thus require $a + b = 0$, and $ab + 2 = -1$, from which we may take without loss of generality that $a = \sqrt{3}$, $b = -\sqrt{3}$.  Thus the desired factorization is $$x^6 + 1 = (x^2 + 1)(x^2 + \sqrt{3}x + 1)(x^2 - \sqrt{3}x + 1).$$  Now we suppose there exist variables $A, B, C, D, E, F$ such that $$\frac{1}{x^6+1} = \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{x^2 + \sqrt{3}x + 1} + \frac{Ex + F}{x^2 - \sqrt{3}x + 1}.$$  Multiplying both sides by $x^6 + 1$ and equating coefficients on the RHS to the LHS (all are zero except the constant coefficient) gives a system of equations in the unknown variables, the solution of which is $$(A,B,C,D,E,F) = \left(0, \tfrac{1}{3}, \tfrac{1}{2\sqrt{3}}, \tfrac{1}{3}, -\tfrac{1}{2\sqrt{3}}, \tfrac{1}{3}\right).$$  This gives us three integrands of the form $$\frac{P(x)}{Q(x)}$$ where $P$ is at most linear, and $Q$ is quadratic, which are easily (but tediously) handled.  For instance, $$\int \frac{\sqrt{3}x + 2}{6(x^2+\sqrt{3}x+1)} \, dx = \frac{\sqrt{3}}{12} \int \frac{2x + \sqrt{3}}{x^2 + \sqrt{3}x + 1} \, dx - \frac{1}{12} \int \frac{dx}{x^2 + \sqrt{3}x + 1},$$ and the first integral is simply a logarithm, and the second is an arctangent after completing the square in the denominator:  $x^2 + \sqrt{3}x + 1 = (x + \tfrac{\sqrt{3}}{2})^2 + (\tfrac{1}{2})^2$.
