How to evaluate $\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}dx$? How to evaluate the following integral
$$I=\int_{0}^{\infty}\dfrac{(x^2-1)\ln{x}}{1+x^4}dx=\dfrac{\pi^2}{4\sqrt{2}}$$
without using residue or complex analysis methods?

Consider the contour integral
$$\oint_C dz \frac{(z^2-1) \log^2{z}}{1+z^4} $$
where $C$ is a keyhole contour about the positive real axis having an outer radius $R$ and an inner radius $\epsilon$.  As $R \to \infty$ and $\epsilon \to 0$, the integral may be shown to be equal to
$$-i 4 \pi \int_0^{\infty} dx \frac{(x^2-1) \log{x}}{1+x^4} + 4 \pi^2 \int_0^{\infty} dx \frac{x^2-1}{1+x^4} $$
The contour integral is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand, which are at $e^{i (2 k+1) \pi/4}$ for $k=0,1,2,3$, or
$$i \frac{\pi}{2} \left [\frac{(i-1) (-\pi^2/16)}{e^{i 3 \pi/4}} - \frac{(i+1) (-9\pi^2/16)}{e^{i  \pi/4}} + \frac{(i-1) (-25 \pi^2/16)}{e^{-i \pi/4}} - \frac{(i+1) (-49\pi^2/16)}{e^{-i 3 \pi/4}}  \right ]$$
which simplifies to $-i (\pi^3/32) 16 \sqrt{2} = -i \pi^3/\sqrt{2}$.  Equating real and imaginary parts, we find that
$$\int_0^{\infty} dx \frac{(x^2-1) \log{x}}{1+x^4} = \frac{\pi^2}{4 \sqrt{2}} $$
