I have stumbled upon the following integral:$$I=\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2} \frac{dx}{\sqrt x}=-\frac{\pi}{24}$$ Although I could solve it, I am not quite comfortable with the way I did it.
But first I will show the way. We can substitute $\ln x \rightarrow t\ $ which gives: $$I=\int_{-\infty}^\infty \frac{t}{\pi^2+t^2}\frac{e^{\frac{t}{2}}}{(1+e^t)^2}dt\overset{t=-x}=\int_{-\infty}^\infty \frac{-x}{\pi^2+x^2}\frac{e^{-\frac{x}{2}}}{(1+e^{-x})^2}dx$$ Also adding the two integral from above and simplify some of it yields: $$2I= \int_{-\infty}^\infty \frac{x}{\pi^2+x^2}\left(\frac{e^{\frac{x}{2}}}{(1+e^x)^2}-\frac{e^{-\frac{x}{2}}}{(1+e^{-x})^2}\right)dx$$ $$\Rightarrow I=-\frac{1}{4} \int_{-\infty}^\infty \frac{x}{\pi^2+x^2}\frac{\sinh \left(\frac{x}{2}\right)}{\cosh ^2\left(\frac{x}{2}\right)}dx$$ And now a round of IBP gives: $$I=\frac12 \int_{-\infty}^\infty \left(\frac{x^2-\pi^2}{(x^2+\pi^2)^2}\right)\left(\frac{1}{\cosh \left(\frac{x}{2}\right)}\right)dx$$ Using the Plancherel theorem the integral simplifies to: $$I=\int_0^\infty \left(\sqrt{\frac{\pi}{2}}x\left(-e^{-\pi x}\right)\right)\left(\sqrt{2\pi}\frac{1}{\cosh(\pi x)}\right)dx\overset{\pi x\rightarrow x}=-\frac{1}{\pi}\int_0^\infty \frac{x}{\cosh( x)}e^{- x}dx$$ We also have the following Laplace tranform for:$$f(t)=\frac{t}{\cosh( t)}\rightarrow F(s)=\frac18\left(\psi_1\left(\frac{s+1}{4}\right)-\psi_1\left(\frac{s+3}{4}\right)\right)$$ Where $\displaystyle{\psi_1(z)=\sum_{n=0}^\infty \frac{1}{(z+n)^2}}\,$ is the trigamma function. $$\Rightarrow I=-\frac{1}{\pi}F(s=1)=-\frac{1}{\pi}\cdot \frac18\left(\psi_1\left(\frac{1}{2}\right)-\psi_1 (1)\right)=-\frac{1}{\pi}\cdot \frac18\left(\frac{\pi^2}{2}-\frac{\pi^2}{6}\right)=-\frac{\pi}{24}$$ Have I done anything wrong, or can it be improved? I have to admit that I mostly used wolfram when applying Plancherel theorem and Laplace transform which I'm not comfortable with, but I didn't find an alternative method myself.
For this question I would like to see a different proof that doesn't rely on that theorem.
Probably not needed, but I should mention that my contour integration knowledge is pretty low. Also maybe there is a conexion with this integral, but I didn't find any.