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.

$$ \begin{align} \int_{0}^{1} \frac{x^{4}+1}{x^{6}+1} \ dx &= \int_{0}^{1} \frac{x^{4}+1+x^{2}-x^{2}}{x^{6}+1} \ dx \\ &= \int_{0}^{1}  \frac{x^{4}-x^{2}+1}{x^{6}+1} \ dx + \int_{0}^{1}\frac{x^{2}}{x^{6}+1} \ dx \\ &= \int_{0}^{1}  \frac{x^{4}-x^{2}+1}{(x^{2}+1)(x^{4}-x^{2}+1)} \ dx + \int_{0}^{1}\frac{x^{2}}{(x^{3})^{2}+1} \ dx \\  &= \int_{0}^{1} \frac{1}{x^{2}+1} \ dx + \frac{1}{3} \int_{0}^{1} \frac{1}{u^{2}+1} \ du \\ &= \frac{4}{3} \int_{0}^{1} \frac{1}{x^{2}+1} \ dx \\&= \frac{4}{3} \left(\frac{\pi}{4} \right) \\ &= \frac{\pi}{3} \end{align}$$
