Solve $\int_{0}^{1}\frac{1}{1+x^6} dx$ Let $$x^3 = \tan y\ \ \text{ so that }\ x^2 = \tan^{2/3}y$$
$$3x^2dx = \sec^2(y)dy$$
$$\int_{0}^{1}\frac{1}{1+x^6}dx = \int_{1}^{\pi/4}\frac{1}{1+\tan^2y}\cdot \frac{\sec^2y}{3\tan^{2/3}y}dy = \frac{1}{3}\int_{1}^{\pi/4} \cot^{2/3}y\ dy$$
How should I proceed after this?
EDITED: Corrected the final integral and the limit from $45$ to $\pi/4$

Continuing where MK12 left off, we proceed as follows:
$$\frac{2-x^2}{x^4-x^2+1} = \frac{1}{2} \times \frac{4-2x^2}{x^4-x^2+1} = \frac{1}{2} \times \frac{1+x^2 + 3(1-x^2)}{x^4-x^2+1}$$
Split the sub-integral into two parts. For one, make the substitution $u=x+\frac{1}{x}$ and the other $v = x-\frac{1}{x}$
Then the rest should be fairly trivial, provided you are meticulously careful with the limits of integration.
