<s>[INST] Integral $\int_0^\infty\frac{dx}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2x^2+2}$ I need your help with this integral:
$$\int_0^\infty\frac{dx}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2\,x^2+2}.$$
I wasn't able to evaluate it in a closed form, although an approximate numerical evaluation suggested its value could be $\frac{\pi}{4}$.
 [/INST] Let us introduce the notation
$$\mathcal{I}=\int_0^{\infty}\frac{dx}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2x^2+2}.$$


*

*Now observe that
\begin{align}\frac{1}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2x^2+2}&=
\frac{1}{x^2+1}\cdot\frac{1}{\frac{x^2-1}{x\cos(\pi\ln x)+1}+2}=\\
&=\frac{1}{x^2+1}\cdot\frac{\cos(\pi\ln x)+\frac1x}{x+\frac1x+2\cos(\pi\ln x)}.
\end{align}

*Using this formula and making the change of variables $x\leftrightarrow \frac1x$, we can rewrite $\mathcal{I}$ as
$$\mathcal{I}=\int_0^{\infty}\frac{1}{x^2+1}\cdot\frac{\cos(\pi\ln x)+x}{x+\frac1x+2\cos(\pi\ln x)}dx.$$

*Summing the last  representation with the initial one, we get
$$2\mathcal{I}=\int_0^{\infty}\frac{dx}{1+x^2}=\frac{\pi}{2}\quad \Longrightarrow\quad \mathcal{I}=\frac{\pi}{4}.$$
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