<s>[INST] Integrate expression with $x^6$ So I'm trying to integrate this expression, but I'm not figuring out what's the best substitution to do... 
$ \int \frac {1}{x^6+1} dx $
I tried to take $x^6 +1 $ and write $ (x^2 + 1) (x^4 -x^2 + 1) $ and then do partial functions, so I reach to the sum of two expressions.
One of them is easy to integrate...
But the other one: 
$ \frac {x^2-1} {x^4 - x^2 + 1} $
I'm having trouble integrating... 
Is there any easier method or is there a way to integrate this right from the beginning or can someone give a hint about how to integrate this last expression? 
Thanks! 
 [/INST] HINT: the partialfrac decomposition is given by $$1/3\, \left( {x}^{2}+1 \right) ^{-1}+1/3\,{\frac {-{x}^{2}+2}{{x}^{4}-
{x}^{2}+1}}
$$
and $$x^4-x^2+1=(x^2-1/2)^2+\frac{3}{4}$$
$$x^4-x^2+1=(x^2+1)^2-3x^2$$ is better (see above) since we get a product finally we obtain
$$1/3\,\arctan \left( x \right) -1/12\,\sqrt {3}\ln  \left( {x}^{2}-
\sqrt {3}x+1 \right) +1/6\,\arctan \left( 2\,x-\sqrt {3} \right) +1/12
\,\sqrt {3}\ln  \left( {x}^{2}+\sqrt {3}x+1 \right) +1/6\,\arctan
 \left( 2\,x+\sqrt {3} \right) 
$$ for the first integral and
$$1/6\,\sqrt {3}\ln  \left( {x}^{2}-\sqrt {3}x+1 \right) -1/6\,\sqrt {3}
\ln  \left( {x}^{2}+\sqrt {3}x+1 \right) 
 $$ for the second one
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