<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] You can take the long way using partial fractions, which gives you:
$$\underbrace{\frac{1}{3(x^2+1)}}_{f_1(x)}+\underbrace{\frac{\sqrt 3 x+2}{6(x^2+\sqrt3 x+1)}}_{f_2(x)}+\underbrace{\frac{3\sqrt3x-6}{18(-x^2+\sqrt3x-1)}}_{f_3(x)}$$
$$\int_0^1f_1(x)dx=\frac 13\int_0^1\frac 1{1+x^2}dx=\left[\arctan(x)\right]_0^1$$
$$\int_0^1f_2(x)dx=\int_0^1\frac{\sqrt 3 x+2}{6(x^2+\sqrt3 x+1)}dx=\frac 16\int_0^1\frac{\sqrt3x+1}{x^2+\sqrt3 x+1}+\frac{1}{x^2+\sqrt3 x+1}$$
From here on out it's pretty straightforward. The answer will involve logarithms and $\arctan$'s. It's similar for $f_3(x)$...
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