<s>[INST] 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?
 [/INST] \begin{align}
&\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}dx 
=2\int_{0}^{1}\frac{(x^2-1)\ln{x}}{1+x^4}dx \\
=&\ \sqrt2 \int_{0}^{1}\ln x \ d\left( -\tanh^{-1}\frac{\sqrt2x}{1+x^2}\right)
\overset{ibp}={\sqrt2} \int_0^1 \frac1x \tanh^{-1}\frac{\sqrt2x}{1+x^2}dx\\
=& \ {\sqrt2} \int_0^1 \int_0^{\pi/4} 
\frac{2(x^2+1)\cos t}{(x^2+1)^2-4x^2\sin^2t}dt\ dx\\ 
= & \ {\sqrt2} \int_0^{\pi/4} \tan^{-1}\frac{2x\cos t}{1-x^2}\bigg|_{x=0}^{x=1}\ dt=  {\sqrt2} \int_0^{\pi/4} \frac\pi2 dt
=\frac{\pi^2}{4\sqrt{2}}
\end{align}
</s>