Integral $\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2} \frac{dx}{\sqrt x}$ 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.

From the identity
$$\Im\int_0^\infty e^{-(\pi-it)x}\,dx=\frac t{\pi^2+t^2}$$
we see that it suffices to compute the imaginary part of the integral
$$\int_0^\infty dx\int_{-\infty}^\infty dt\;
\frac{e^{\alpha t}}{(1+e^t)^2}e^{-\pi x}$$
where $\alpha=1/2+ix$. Now, the integral with respect to $t$ is easy by means of the substitution $u=e^t$ and using the beta function. We thus get
$$\int_0^\infty \pi\left(\frac12-ix\right)\frac{e^{-\pi x}}{\cosh(\pi x)}\,dx.$$
Taking the imaginary part we see that the problem boils down to compute the integral
$$\int_0^\infty \frac{x e^{-\pi x}}{\cosh(\pi x)}\,dx
=2\int_0^\infty \frac{x}{1+e^{2\pi x}}\,dx$$
which, after the substitution $v=2\pi x$, reduces to the integral representation of the eta function $\eta(2)$. Also notice that taking the real part we obtain the evaluation 
$$\int_0^\infty\frac1{(\pi^2+\log^2 x)(1+x)^2} \frac{dx}{\sqrt x}=
\frac{\log2}{2\pi}.$$
This method generalises to other integrals such as
\begin{align*}
\int_0^\infty \frac{1}{(\pi^2+\ln^2 x)(1+x)^3} \frac{dx}{\sqrt x}
&=\frac{3\log (2)}{8 \pi }-\frac{3 \zeta (3)}{16 \pi ^3}\\
\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^3} \frac{dx}{\sqrt x}
&=-\frac{\pi }{24}\\
\int_0^\infty \frac{1}{(\pi^2+\ln^2 x)(1+x)^4} \frac{dx}{\sqrt x}
&=\frac{5 \log (2)}{16 \pi }-\frac{9 \zeta (3)}{32 \pi ^3}\\
\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^4} \frac{dx}{\sqrt x}
&=-\frac{223 \pi }{5760}\\
\int_0^\infty \frac{1}{(\pi^2+\ln^2 x)(1+x)^5} \frac{dx}{\sqrt x}
&=-\frac{43 \zeta (3)}{128 \pi ^3}+\frac{15 \zeta (5)}{256 \pi ^5}+\frac{35 \log (2)}{128
   \pi }\\
\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^5} \frac{dx}{\sqrt x}
&=-\frac{103 \pi }{2880}\\
\end{align*}
Incidentally, since the integrals $\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^k} \frac{dx}{\sqrt x}$ both yield the same value for $k=2,3$, we also deduce
$$\int_0^\infty \frac{\sqrt x\ln x}{(\pi^2+\ln^2 x)(1+x)^3}\;dx=0.$$
This, however, should not be surprising due to the symmetry $x\mapsto1/x$. 
