<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] The integrand is invariant under inversion:
$$\log^2(1/x){1+(1/x)^2\over1+(1/x)^4}d(1/x)=-\log^2x{1+x^2\over1+x^4}dx$$
Thus
$$\begin{align}
\int_0^\infty \log^2x{1+x^2\over1+x^4}dx&=\int_0^1 \log^2x{1+x^2\over1+x^4}dx+\int_1^\infty \log^2x{1+x^2\over1+x^4}dx\\
&=\int_0^1 \log^2x{1+x^2\over1+x^4}dx-\int_1^0 \log^2x{1+x^2\over1+x^4}dx\\
&=2\int_0^1 \log^2x{1+x^2\over1+x^4}dx
\end{align}$$
Maye now you can expand things as a power series.
</s>