Compute: $\int_{0}^{1}\frac{x^4+1}{x^6+1} dx$ I'm trying to compute: $$\int_{0}^{1}\frac{x^4+1}{x^6+1}dx.$$
I tried to change $x^4$ into $t^2$ or $t$, but it didn't work for me.
Any suggestions?
Thanks!

one way is partial fractions on
$$
\frac{x^4+1}{x^6+1}=\frac{(x-e^{\pi i/4})(x-e^{3\pi i/4})(x-e^{5\pi i/4})(x-e^{7\pi i/4})}{(x-e^{\pi i/6})(x-e^{3\pi i/6})(x-e^{5\pi i/6})(x-e^{7\pi i/6})(x-e^{9\pi i/6})(x-e^{11\pi i/6})}
$$
$$
=\frac{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}{(x^2+1)(x^2+\sqrt{3}x+1)(x^2-\sqrt{3}x+1)}
$$
