<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] Following Mhenni's suggestion, I will calculate $$I(\mu) = \int_0^{\infty} x^{\mu}\frac{1+x^2}{1+x^4} \,dx $$ and then take $I''(0)$.
By the ubiquitous formula $$\int_0^{\infty}\frac{x^a}{1+x^b} \,dx =\frac{\pi}{b \sin(\pi(a+1)/b)}$$
we obtain $$I(\mu)=\frac{\pi}{4}   \left[ \frac{1}{\sin(\pi(\mu+1)/4)} + \frac{1}{\sin(\pi(\mu+3)/4)}      \right]$$
This gives (after some simple algebra)
$$I''(0) = \int_0^{\infty}   \frac{1+x^2}{1+x^4} \log^2 x \,dx  = \frac{\pi}{4}\left[\frac{3 \pi ^2}{8 \sqrt{2}}+\frac{3 \pi ^2}{8 \sqrt{2}} \right] = \frac{3 \pi ^3}{16 \sqrt{2}}$$
as was to be shown.
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