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.

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 $s=2$.
