Conjectured value of $\int_{0}^{\infty}\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln x}\right)\frac{\mathrm{d}x}{x^2+1}$ I was curious whether this integral has a closed form expression :

$$\int_{0}^{\infty}\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln
 x}\right)\frac{\mathrm{d}x}{x^2+1}$$

The integrand has a singularity at $x=1$, but it's removable.  And as $x \to \infty$, the integrand behaves like $\frac{1}{x \ln^{2}x}$. So the integral clearly converges.
Although I have not been able to derive its closed form, I think, by reverse symbolic calculators, up to 20 digits it could be
$$I=\frac{4G}{\pi}$$
where $G$ is Catalan's constant. Is it true or is it completely fabulous?
EDIT. NOTE :
For better search to this integral I have renamed the title from Conjectured value of logarithmic definite integral, which is ambiguous and did not say anything, to the current one with integral explicitly written.
 A: Jack D'Aurizio showed  that $$\int_{0}^{\infty}\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln
 x}\right)\frac{dx}{x^2+1} = \int_{0}^{\infty} \left( 1-\frac{1}{\cosh x} \right) \frac{dx}{x^{2}} .$$
The following is an alternative evaluation of the integral on the right.

An integral representation of the Dirichlet beta function is $$\beta(s) = \frac{1}{ 2 \, \Gamma(s)} \int_{0}^{\infty} \frac{x^{s-1}}{\cosh(x)} \, dx \, , \quad \text{Re}(s) >0\tag{1}. $$
And the Laplace transform of $x^{s-1}$ is $$\int_{0}^{\infty} x^{s-1} e^{-ax} \, dx = \frac{\Gamma(s)}{a^{s}} \, , \quad (\text{Re}(s) >0, \ \text{Re}(a)>0) \tag{2}.$$
Subtracting $(1)$ from $(2)$, we get $$\int_{0}^{\infty}\left(e^{-ax} - \frac{1}{\cosh (x)} \right) x^{s-1} \, dx = \Gamma(s) \left( a^{-s}   - 2  \beta(s) \right)  ,  \tag{3}$$ which holds for $ \text{Re}(s) > -1$ and $\text{Re}(a) > 0$.
If we restrict $s$ to that strip $-1 < \text{Re}(s) <0$, then $(3)$ also holds for $a = 0$.
From the functional equation of the Dirichlet beta function, we see that the Dirichlet beta function has a zero at $s=-1$.
So letting $s$ tend to $-1$, we get
$$ \begin{align} \int_{0}^{\infty} \left(1- \frac{1}{\cosh x} \right) \frac{dx}{x^{2}}  &= \lim_{s \downarrow -1} \Gamma(s) \left((0 - 2 \beta(s)\right) \\ &= - 2 \lim_{s \downarrow -1} \Gamma(s) \beta(s) \\ &=-2 \lim_{s \downarrow -1} \left(-\frac{1}{s+1} + \mathcal{O}(1) \right) \beta(s) \\ &= 2 \beta'(-1). \end{align}$$
To show that $ \displaystyle \beta'(-1) = \frac{2G}{\pi} $, differentiate both sides of the functional equation, and then let $s=2$.
A: It is not necessary to exploit any symmetries of the integrand.
Setting $x=e^y$
$$
I=\int_{-\infty}^{\infty}\underbrace{e^y\left(\frac{e^y-1}{y^2}-\frac{1}{y}\right)\frac{1}{e^{2y}+1}}_{f(y)}\,dy
$$
Integrating around a big semicircle in the UHP (exercise: show convergence in this domain of the complex plane) we obtain
$$
I=2 \pi i \sum_{n=0}^{\infty}\text{Res}(f(z),z=z_n)
$$
here $z_n=\frac{i\pi}2(2n+1)$. This is easily rewritten as
$$
I=2 \pi i\left(\left(\frac{1}{\pi}\color{blue}{\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}}-\frac{2}{\pi^2}\color{red}{\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}}\right) -\frac{2i}{\pi^2}\color{green}{\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}}\right)
$$
since $\color{blue}{\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}}$ and $\color{red}{\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}}$ the imaginary parts cancel and we are left with

$$
I= \frac{4}{\pi}\color{green}{\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}}=\frac{4\color{green}{K}}{\pi}
$$

A: Our integral equals
$$ I=\int_{-\infty}^{+\infty}\left(\frac{e^t-1-t}{t^2}\right)\frac{e^t}{e^{2t}+1}\,dt $$
that by exploiting symmetry becomes
$$ I = \int_{0}^{+\infty}\frac{e^{t}+e^{-t}-2}{t^2(e^{t}+e^{-t})}\,dt =\int_{0}^{+\infty}\frac{\cosh(t)-1}{t^2\cosh(t)}\,dt$$
The last integral is straightforward to compute trough the residue theorem. Since
$$ \text{Res}\left(\frac{\cosh(t)-1}{t^2\cosh(t)},t=\frac{\pi(2k+1)}{2}i\right)= (-1)^{k+1}\frac{4i}{\pi^2(2k+1)^2}$$
we have:
$$\boxed{ I = \frac{4}{\pi}\sum_{k\geq 0}\frac{(-1)^k}{(2k+1)^2}=\color{red}{\frac{4G}{\pi}}}$$
as conjectured.
A: Here is yet another approach.  We first note that we can write $\frac{x-1}{\log(x)}$ as 
$$\frac{x-1}{\log(x)}=\int_0^1 x^t\,dt$$
Therefore, we can write
$$\begin{align}
\int_0^\infty \left(\frac{x-1}{\log^2(x)}-\frac{1}{\log(x)}\right)\frac{1}{1+x^2}\,dx&=\int_0^\infty \int_0^1 \frac{x^t-1}{\log(x)}\,\frac{1}{1+x^2}\,dt\,dx\\\\
&=\int_0^1 \int_0^\infty \frac{x^t-1}{\log(x)}\,\frac{1}{1+x^2}\,dx\,dt\tag1
\end{align}$$
Let $I(t)$ represent the inner integral of the right-hand side of $(1)$.  Then, differentiating, we find that
$$\begin{align}
I'(t)&=\int_0^\infty \frac{x^t}{1+x^2}\,dx\\\\
&=\frac{\pi}{2\cos(\pi t/2)}\tag 2
\end{align}$$
where I derived the right-hand side of $(2)$ in THIS ANSWER.  Alternatively, using real analysis only, we have 
$$\begin{align}
\int_0^\infty \frac{x^t}{1+x^2}\,dx&=\frac12 B\left(\frac{1+t}{2},\frac{1-t}{2}\right)\\\\
&=\frac12 \Gamma\left(\frac{1+t}{2}\right)\Gamma\left(\frac{1-t}{2}\right)\\\\
&=\frac12\frac{\pi}{\sin\left(\pi\frac{1+t}{2}\right)}\\\\
&=\frac{\pi}{2\cos(\pi t/2)}
\end{align}$$
Integrating $(2)$ and using $I(0)=0$ reveals
$$I(t)=\int_0^t \frac{\pi}{2\cos(\pi t'/2)}\,dt' \tag 3$$
Substituting $(3)$ into $(1)$ yields
$$\begin{align}
\int_0^\infty \left(\frac{x-1}{\log^2(x)}-\frac{1}{\log(x)}\right)\frac{1}{1+x^2}\,dx&=\frac{\pi}{2}\int_0^1 \int_0^t \sec(\pi t'/2)\,dt'\,dt \tag 4\\\\
&=\frac{\pi}{2}\int_0^1 (1-t)\sec(\pi t/2)\,dt \tag5\\\\
&=\frac{\pi}{2}\int_0^1 t\csc(\pi t/2)\,dt \tag 6\\\\
&=\frac{1}{\pi}\int_{-\pi/2}^{\pi/2}\frac{t}{\sin(t)}\,dt \tag 7\\\\
&=\frac{4G}{\pi} \tag 8
\end{align}$$
as was to be shown!

NOTES:
In going from $(4)$ to $(5)$, we changed the order of integration and carried out the inner integral.
In going from $(5)$ to $(6)$, we enforced the substitution $t \to 1-t$.
In going from $(6)$ to $(7)$, we enforced the substitution $t \to 2t/\pi$ and exploited the evenness of the integrand.
In going from $(7)$ to $(8)$, we made use of one of the integral identities for Catalan's Constant as found HERE.

ALTERNATIVE DEVELOPMENT
Note that we can write $(3)$ as 
$$I(t)=\log\left(\cot\left(\frac{\pi}{4}(1-t)\right)\right) \tag 9$$
Then, substituting $(9)$ into $(1)$ yields
$$\begin{align}
\int_0^\infty \left(\frac{x-1}{\log^2(x)}-\frac{1}{\log(x)}\right)\frac{1}{1+x^2}\,dx&=\int_0^1 \log\left(\cot\left(\frac{\pi}{4}(1-t)\right)\right)\,dt \\\\
&=\frac{4}{\pi}\int_0^{\pi/4} \log(\cot(t))\,dt \tag 9\\\\
&=\frac{4G}{\pi} 
\end{align}$$
which uses another well-known integral identity for $G$ as found HERE.
Note that if we enforce the substitution $t\to \text{arccot}(t)$ in $(9)$, we find the result in terms of the series representation of $G$ as
$$\begin{align}
\int_0^\infty \left(\frac{x-1}{\log^2(x)}-\frac{1}{\log(x)}\right)\frac{1}{1+x^2}\,dx&=\frac{4}{\pi}\int_0^{\pi/4} \log(\cot(t))\,dt \\\\
&=\frac{4}{\pi}\int_1^{\infty}\frac{\log(t)}{1+t^2}\,dt\\\\
&=-\frac{4}{\pi}\int_0^1 \frac{\log(t)}{1+t^2}\,dt\\\\
&=-\frac{4}{\pi}\sum_{n=0}^\infty(-1)^n \int_0^1 t^{2n}\log(t)\,dt\\\\
&=\frac{4}{\pi}\sum_{n=0}^\infty(-1)^n \int_0^1 \frac{t^{2n}}{2n+1}\,dt\\\\
&=\frac{4}{\pi}\sum_{n=0}^\infty(-1)^n \frac{1}{(2n+1)^2}\\\\
&=\frac{4G}{\pi}
\end{align}$$
as expected once again!
A: Though using the residue method is somewhat straightforward, but not everyone can understand it. So, here is a residue-free method:
Split the integral into two terms where each term is in the interval $0<x<1$ and $1<x<\infty$, then use the substitution $x\mapsto\frac{1}{x}$ to the second term. We will get
$$
\left[\int_{0}^{1}+\int_{1}^{\infty}\right]\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln x}\right)\frac{\mathrm{d}x}{x^2+1}=\int_{0}^{1}\frac{(x-1)^2}{x\ln^2 x}\cdot\frac{\mathrm{d}x}{x^2+1}\tag1
$$
Now, for $a\ge-1$ , one may consider the following integral
$$
I(a)=\int_{0}^{1}x^a\cdot\frac{(x-1)^2}{\ln^2 x}\cdot\frac{\mathrm{d}x}{1+x^2}\tag2
$$
and the desired integral is $I(-1)$. Since $0<x<1$, one may observe that $I(\infty)\to0$ as $a\to\infty$.
\begin{align}
I''(a)&=\int_{0}^{1}\frac{x^a(x-1)^2}{1+x^2}\ \mathrm{d}x\\[10pt]
&=\int_{0}^{1}\sum_{k=0}^\infty(-1)^k\ x^{2k+a}\ (x^2-2x+1)\ \mathrm{d}x\\[10pt]
&=\sum_{k=0}^\infty(-1)^k\left(\frac{1}{2k+a+3}-\frac{2}{2k+a+2}+\frac{1}{2k+a+1}\right)\\[10pt]
&=\frac{1}{4}\left[\psi\left(\frac{a+5}{4}\right)-2\psi\left(\frac{a+4}{4}\right)+2\psi\left(\frac{a+2}{4}\right)-\psi\left(\frac{a+1}{4}\right)\right]\\[10pt]
I'(a)&=\ln\Gamma\left(\frac{a+5}{4}\right)-\ln\Gamma\left(\frac{a+1}{4}\right)+2\ln\Gamma\left(\frac{a+2}{4}\right)-2\ln\Gamma\left(\frac{a+4}{4}\right)\\[10pt]
I(a)&=4\left[\psi\left(-2,\frac{a+5}{4}\right)-\psi\left(-2,\frac{a+1}{4}\right)+2\psi\left(-2,\frac{a+2}{4}\right)-2\psi\left(-2,\frac{a+4}{4}\right)\right]\tag3\\[10pt]
\end{align}
Hence
$$
I(-1)=4\left[\psi\left(-2,1\right)-\psi\left(-2,0\right)+2\psi\left(-2,\frac{1}{4}\right)-2\psi\left(-2,\frac{3}{4}\right)\right]=\frac{4G}{\pi}
$$
Wolfram Alpha confirms it. One may also use the special values of generalized polygamma function and its related relation with derivative of Hurwitz Zeta Function: $$\psi(-2,x)=\zeta'(-1,x)-\frac{x^2}{2}+\frac{x}{2}-\frac{1}{12}$$
