Calculate $\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$ I have to calculate the integral
$$\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$$
I've calculated the integral $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x= \sqrt{2}  \pi $. Then $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x=2 \int_{0}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x$. How do I now arrive at the initial integral?

Note
$$\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} dx
= \int_{0}^{1} \frac{1+\frac1{x^{2}}}{x^{2}+\frac1{x^2}} dx
=\int_0^1 \frac {d(x-\frac1x)}{(x-\frac1x)^2+2}
= \frac\pi{2\sqrt2}
$$
