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

$$\begin{align}
\int_0^1 \frac{1+x^2}{1+x^4} \operatorname d \!x &= 
\int_0^1  \frac{1+x^4-x^4+x^2}{1+x^4} \operatorname d \!x\\
&= 1 + \int_0^1 x^2\frac{1-x^2}{1+x^4} \operatorname d \!x\\
&= 1 + \int_0^1 x^2\frac{1 + x^4 - x^4-x^2}{1+x^4} \operatorname d \!x\\
&= 1 + \frac{1}{3} -\int_0^1 x^4\frac{1+x^2}{1+x^4} \operatorname d \!x\ \ldots
\end{align}
$$
and so forth.
