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.

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: $$\psi(-2,x)=\zeta'(-1,x)-\frac{x^2}{2}+\frac{x}{2}-\frac{1}{12}$$
