<s>[INST] 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 $\pi/4$
 [/INST] Hint:
note that 
$$
x^6+1=(x^2+1)(x^2-\sqrt{3}+1)(x^2+\sqrt{3}+1)
$$
than use partial fraction decomposition.
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