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

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 $x\mapsto1/x$.

That $\pi^2+\log^2(x)$ makes me think to the integral representation for Gregory coefficients:

$$ \int_{0}^{+\infty}\frac{dx}{(1+x)^n (\pi^2+\log^2 x)} = \frac{1}{n!}\left[\frac{d^n}{dx^n}\frac{z}{\log(1-z)}\right]_{z=0}=[z^n]\frac{z}{\log(1-z)} \tag{1}$$ which can be seen as a consequence of the Lagrange-Buhrmann inversion theorem. We just need to insert a factor $\frac{\log x}{\sqrt{x}}$ in the integrand function appearing in the LHS, so let's go back to the residue theorem.

$$ \int_{0}^{+\infty}\frac{\log(x)\,dx}{\sqrt{x}(1+x)^2(\pi^2+\log^2 x)}=\int_{-\infty}^{+\infty}\frac{t e^{-t/2}}{(2\cosh\frac{t}{2})^2 (\pi^2+t^2)}\,dt$$ equals $$ -\frac{1}{4}\int_{\mathbb{R}}\frac{t\sinh\frac{t}{2}}{(t^2+\pi^2)\cosh^2\frac{t}{2}}\,dt=-\int_{\mathbb{R}}\frac{t\sinh t}{(4t^2+\pi^2)\cosh^2 t}\,dt. $$ The meromorphic function $\frac{\sinh t}{\cosh^2 t}=-\frac{d}{dt}\left(\frac{1}{\cosh t}\right)$ only has double poles with residue zero, hence all the mass of the last integral comes from the singularity at $\frac{\pi i}{2}$ and from the behaviour at infinity. The residue theorem grants $$ \frac{1}{\cosh x}=\sum_{n\geq 0}(-1)^n \frac{\pi(2n+1)}{\frac{\pi^2}{4}(2n+1)^2+x^2} $$ and $$ \frac{\sinh x}{\cosh^2 x} = \sum_{n\geq 0}(-1)^n \frac{2\pi(2n+1)x}{(\frac{\pi^2}{4}(2n+1)^2+x^2)^2}.$$ Since $$ \int_{\mathbb{R}}\frac{2\pi(2n+1)x^2}{(\frac{\pi^2}{4}(2n+1)^2+x^2)^2 (\pi^2+4x^2)}\,dx = \frac{1}{2\pi(n+1)^2}$$ our integral equals $-\frac{1}{2\pi}\eta(2)=\color{red}{-\frac{\pi}{24}}$ by the dominated convergence theorem, allowing to switch $\int_{\mathbb{R}}$ and $\sum_{n\geq 0}$. $\frac{1}{24}$ is also the coefficient of $z^3$ in $\frac{z}{\log(1-z)}$, but so far I have not found a direct way to relate the original integral to the $n=3$ instance of $(1)$.

This is not a full answer, but one does not need the full Laplace transform of $x/\cosh x$, as the integral can be done by expanding $1/\cosh x$ into a geometric series

$$\begin{aligned} I = \int_{0}^{\infty}\frac{xe^{-x}}{\cosh x}\,\mathrm{d}x &= 2\int_{0}^{\infty}\frac{xe^{-x}}{e^{x}}\frac{\mathrm{d}x}{1+e^{-2x}} = 2\sum_{n=0}^{\infty}(-1)^{n}\int_{0}^{\infty}xe^{-(2+2n)x}\,\mathrm{d}x \\ &= 2\sum_{n=0}^{\infty}\frac{(-1)^{n}}{4(1+n)^{2}}\int_{0}^{\infty}ue^{-u}\,\mathrm{d}u = \frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(1+n)^{2}} \\ &= \frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{2}} = \frac{\eta(2)}{2} = \frac{\pi^{2}}{24}\end{aligned}$$

where $\eta(s)$ is the Dirichlet eta function.