Integral $\int\frac{4x^4}{{x^8+1}}\;dx$ I'm attempting this integral but I'm unsure how to proceed. I've begun to suspect it's actually non-elementary. Can anyone do this? I could always use Taylor series after a small bit of u-sub and integration by parts, but I wanted to know if there were a somewhat more direct way of doing it.
 The integral is $$\int\frac{4x^4}{{x^8+1}}\;dx$$
Thanks in advance.

First of all, let me precise that, for this kind of integrals, Taylor expansions could me more thatn dangerous.
Considering the integrand, you can first write (by analogy with $(x^4+1)$
$$\frac {4x^4}{x^8+1}=\frac{\sqrt{2} x^2}{x^4-\sqrt{2} x^2+1}-\frac{\sqrt{2} x^2}{x^4+\sqrt{2} x^2+1}$$
Then
$$x^4-\sqrt{2} x^2+1=\left(x^2-\frac{1+i}{\sqrt{2}}\right) \left(x^2-\frac{1-i}{\sqrt{2}}\right)$$
$$x^4+\sqrt{2} x^2+1=\left(x^2+\frac{1-i}{\sqrt{2}}\right) \left(x^2+\frac{1+i}{\sqrt{2}}\right)$$ which finally make
$$\frac {4x^4}{x^8+1}=\frac{i \sqrt{2}}{\sqrt{2}-(1-i) x^2}-\frac{i \sqrt{2}}{\sqrt{2}-(1+i) x^2}+\frac{i \sqrt{2}}{\sqrt{2}+(1-i) x^2}-\frac{i \sqrt{2}}{\sqrt{2}+(1+i) x^2}$$ and now, we face quite trivial integrals.
$$\int\frac {4x^4}{x^8+1}\,dx=\cos \left(\frac{\pi }{8}\right) \left(\frac{1}{2} \log \left(\frac{x^2+2 x \sin
   \left(\frac{\pi }{8}\right)+1}{x^2-2 x \sin \left(\frac{\pi
   }{8}\right)+1}\right)+\tan ^{-1}\left(\frac{2 x \sin \left(\frac{\pi
   }{8}\right)}{1-x^2}\right)\right)+$$
$$\sin \left(\frac{\pi }{8}\right) \left(\frac{1}{2} \log \left(\frac{x^2-2 x \cos
   \left(\frac{\pi }{8}\right)+1}{x^2+2 x \cos \left(\frac{\pi
   }{8}\right)+1}\right)-\tan ^{-1}\left(\frac{2 x \cos \left(\frac{\pi
   }{8}\right)}{1-x^2}\right)\right)$$
In particular
$$\int_0^\infty\frac{4x^4}{{x^8+1}}\;dx=\pi\sqrt{1-\frac{1}{\sqrt{2}}} $$
