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 $$\int_0^1 \! \frac{1+x^2}{1+x^4} \, \operatorname d \!x=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\cdots$$

Hints: Expand the denominator as a geometric series in $-x^4$:
$$\frac{1}{1+x^4}=\sum_{n=0}^{\infty}(-1)^nx^{4n}=1-x^4+x^8-x^{12}+...$$
Multiply by $1+x^2$ to obtain a series form for the integrand.
$$\frac{1+x^2}{1+x^4}=\frac{1}{1+x^4}+\frac{x^2}{1+x^4}=1+x^2-x^4-x^6+x^8+x^{10}-x^{12}-x^{14}+...$$
All that's left is to integrate term-by-term.
