How to evaluate $\int_0^1\frac{1+x^4}{1+x^6}\,dx$ $$\int_0^1\frac{1+x^4}{1+x^6}\,dx$$
Can anyone help me solve the question? I am struggling with this.
 A: RV's great answer notwithstanding, here's another way using the residue theorem.  Note that
$$\int_0^1 dx \frac{1+x^4}{1+x^6} = \frac14 \int_{-\infty}^{\infty} dx \frac{1+x^4}{1+x^6}$$
The integral on the RHS is, by the residue theorem, $i 2 \pi$ times the sum of the residues of the poles of the integrand in the upper half plane, or
$$\begin{align}i \frac{\pi}{2} \left (\frac{1+e^{i 4 \pi/6}}{6 e^{i 5 \pi/6}}  + \frac{2}{6 i}+\frac{1+e^{i 20 \pi/6}}{6 e^{i 25 \pi/6}}\right ) &= i\frac{\pi}{6} \left (-i + e^{-i \pi/6} + e^{-i 5 \pi/6} \right )\\ &= \frac{\pi}{6} \left ( 1+2 \sin{\frac{\pi}{6}}\right ) \\ &= \frac{\pi}{3}\end{align}$$
A: $$ \begin{align} \int_{0}^{1} \frac{x^{4}+1}{x^{6}+1} \ dx &= \int_{0}^{1} \frac{x^{4}+1+x^{2}-x^{2}}{x^{6}+1} \ dx \\ &= \int_{0}^{1}  \frac{x^{4}-x^{2}+1}{x^{6}+1} \ dx + \int_{0}^{1}\frac{x^{2}}{x^{6}+1} \ dx \\ &= \int_{0}^{1}  \frac{x^{4}-x^{2}+1}{(x^{2}+1)(x^{4}-x^{2}+1)} \ dx + \int_{0}^{1}\frac{x^{2}}{(x^{3})^{2}+1} \ dx \\  &= \int_{0}^{1} \frac{1}{x^{2}+1} \ dx + \frac{1}{3} \int_{0}^{1} \frac{1}{u^{2}+1} \ du \\ &= \frac{4}{3} \int_{0}^{1} \frac{1}{x^{2}+1} \ dx \\&= \frac{4}{3} \left(\frac{\pi}{4} \right) \\ &= \frac{\pi}{3} \end{align}$$
