Show that $\int_0^1 \! \frac{1+x^2}{1+x^4} \, \operatorname d \!x=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\cdots$ As the title states, trying to solve $$\int_0^1 \! \frac{1+x^2}{1+x^4} \, \operatorname d \!x=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\cdots$$

Simply express the integrand as the series that converges for $|x| < 1$ and then interchange the order of summation and integration.
Hint: It is straightforward to show that
$$\frac{1 + x^2}{1 + x^4} = 1 + x^2 - x^4 - x^6 + x^8 + x^{10} - \cdots$$
because
$$\frac{1 + x^2}{1 + x^4} = \frac{1}{1 + x^4} + \frac{x^2}{1 + x^4}.$$
Hopefully you know what the geometric series are. Now integrate the series above term-by-term.
