<s>[INST] Integral $\int_0^\infty \log^2 x\frac{1+x^2}{1+x^4}dx=\frac{3 \pi^3}{16\sqrt 2}$ This integral below
$$
I:=\int_0^\infty \log^2 x\frac{1+x^2}{1+x^4}dx=\frac{3 \pi^3}{16 \sqrt 2}
$$
is what I am trying to prove.  Thanks.
We can not expand the denominator as a series since the domain of integration is for $x\in [0,\infty)$.  Next I wrote
$$
I=\int_0^\infty \log^2 x \frac{1+x^4-x^4+x^2}{1+x^4}dx=\int_0^\infty \log^2x \left(\frac{1+x^4}{1+x^4}+\frac{x^2-x^4}{1+x^4}\right)dx=\\
\int_0^\infty \log^2 x \, dx+\int_0^\infty \log^2 x \frac{x^2}{1+x^4}dx-\int_0^\infty \log^2 x \frac{x^4}{1+x^4}dx,
$$
however only the middle integral is convergent.  I am not sure how to go about solving this problem.  Thank you
 [/INST] Here is another way to calculate. It is much simpler. Clearly
\begin{eqnarray*}
I&=&2\int_0^1 \log^2 x\frac{1+x^2}{1+x^4}dx\\
&=&2\int_0^1\sum_{n=0}^\infty(1+x^2)(-1)^nx^{4n}\log^2xdx\\
&=&2\int_0^1\sum_{n=0}^\infty(-1)^n(x^{4n}+x^{4n+2})\log^2xdx\\
&=&2\sum_{n=0}^\infty\int_0^1(-1)^n(x^{4n}+x^{4n+2})\log^2xdx\\
&=&4\sum_{n=0}^\infty(-1)^n(\frac{1}{(4n+1)^3}+\frac{1}{(4n+3)^3})\\
&=&4\sum_{n=-\infty}^\infty(-1)^n\frac{1}{(4n+1)^3}\\
&=&\frac{3\pi^3}{16\sqrt2}
\end{eqnarray*}
Here we use the following theorem
$$ \sum_{n=-\infty}^\infty (-1)^nf(n)=-\pi \sum_{k=1}^m\text{Re}(\frac{f(z)}{\sin\pi z},a_k) $$
where $a_1,a_2,\cdots,a_m$ are poles of $f(z)$. For $f(z)=\frac{1}{(4z+1)^3}$, $z=-\frac{1}{4}$ is the only pole and
$$ \text{Re}(\frac{f(z)}{\sin\pi z},-\frac{1}{4})=-\frac{3\pi^2}{64\sqrt2}. $$
Thus we have the result.
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