<s>[INST] 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.
 [/INST] 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: $$\psi(-2,x)=\zeta'(-1,x)-\frac{x^2}{2}+\frac{x}{2}-\frac{1}{12}$$
</s>