How do I find the integral of 1/(x^6+1) using partial fractions?

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Homework Statement



basically the title

Homework Equations





The Attempt at a Solution


so I tried writing it as a difference of squares and got (x^3+1+sqrt(2)*x^1.5)(x^3+1-sqrt(2)x^1.5)
and I attempted partial fractions and I don't know if I did anything wrong, but then I got stuck when it came time to solve for the variables in the partial fraction decomposition. I'm not lost on this problem so If anyone has any clue, please guide me in the right direction. Thanks!
 
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I am not good at integrations... but here is a answer kind of thing done in mathematica...
you can check your results with it...
Sorry, could not really help you.
 

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Try writing 1 + x^6 as a sum of two cubes:
<br /> 1+x^6 = 1 + \left(x^2\right)^3<br />

and factor, then apply partial fractions.
 
Hint :: ##\displaystyle \int\frac{1}{1+x^6}dx = \frac{1}{2}\int\frac{(1+x^4)+(1-x^4)}{1+x^6}dx##

and Break into two parts ##I## and ##J##
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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