How to find ${\large\int}_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx$? Please help me to find a closed form for this integral:
$$
I=\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx
$$
Routine textbook methods for this complicated integral fail. 
 A: Here is another Feynman's way to evaluate the integral. Set $x=\frac1t$, then
$$
\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x}\ dx=\int_0^1\frac{t-1-t\ln t}{(1+t^2)\ln^2t}\ dt.
$$
Now consider
$$
\mathcal{I}(\alpha)=\int_0^1t^\alpha\cdot\frac{t-1-t\ln t}{(1+t^2)\ln^2t}\ dt.
$$
Hence
\begin{align}
\frac{d^2\mathcal{I}}{d\alpha^2}&=\int_0^1\frac{t^\alpha}{1+t^2}(t-1-t\ln t)\ dt\\
&=\int_0^1\sum_{n=0}^\infty(-1)^nt^{2n+\alpha}(t-1-t\ln t)\ dt\\
&=\sum_{n=0}^\infty(-1)^n\int_0^1\left(t^{2n+\alpha+1}-t^{2n+\alpha}-t^{2n+\alpha+1}\ln t\right)\ dt\tag1\\
&=\sum_{n=0}^\infty(-1)^n\left[\frac{1}{2n+\alpha+2}-\frac{1}{2n+\alpha+1}+\frac{1}{(2n+\alpha+2)^2}\right]\tag2\\
&=-\frac14\left[\psi\left(\frac{\alpha+2}{4}\right)-\psi\left(\frac{\alpha+4}{4}\right)\right]+\frac14\left[\psi\left(\frac{\alpha+1}{4}\right)-\psi\left(\frac{\alpha+3}{4}\right)\right]\\
&\quad+\frac1{16}\left[\psi_1\left(\frac{\alpha+2}{4}\right)-\psi_1\left(\frac{\alpha+4}{4}\right)\right]\tag3\\
\frac{d\mathcal{I}}{d\alpha}&=-\ln\Gamma\left(\frac{\alpha+2}{4}\right)+\ln\Gamma\left(\frac{\alpha+4}{4}\right)+\ln\Gamma\left(\frac{\alpha+1}{4}\right)-\ln\Gamma\left(\frac{\alpha+3}{4}\right)\\
&\quad+\frac14\left[\psi\left(\frac{\alpha+2}{4}\right)-\psi\left(\frac{\alpha+4}{4}\right)\right]\tag4\\
&=\ln\Gamma\left[\frac{(\alpha+1)(\alpha+4)}{(\alpha+2)(\alpha+3)}\right]+\frac14\left[\psi\left(\frac{\alpha+2}{4}\right)-\psi\left(\frac{\alpha+4}{4}\right)\right]\tag5\\
\mathcal{I}(\alpha)&=\int\ln\Gamma\left[\frac{(\alpha+1)(\alpha+4)}{(\alpha+2)(\alpha+3)}\right]\ d\alpha+\ln\Gamma\left(\frac{\alpha+2}{4}\right)-\ln\Gamma\left(\frac{\alpha+4}{4}\right).
\end{align}
Since $0<t<1$, then as $\alpha\to\infty$, implying $\mathcal{I}(\alpha)\to0$ and $\mathcal{I}'(\alpha)\to0$. Thus

$$
\mathcal{I}(0)=\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x}\ dx=\color{blue}{\frac{\ln2}6+\frac{\ln\pi}2-6\ln A+\frac2\pi G},
$$

where $A$ is the Glaisher–Kinkelin constant and $G$ is the Catalan constant.

Notes :
$\displaystyle(1)\ \ \int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\  n=0,1,2,\ldots$
$\displaystyle(2)\ \ \sum_{k=0}^\infty\frac{(-1)^{k}}{(z+k)^{m+1}}=\frac1{(-2)^{m+1}m!}\left[\psi_{m}\left(\frac{z}{2}\right)-\psi_{m}\left(\frac{z+1}{2}\right)\right]$
$\displaystyle(3)\ \ \psi_{m}(z) := \frac{d^m}{dz^m} \psi(z) = \frac{d^{m+1}}{dz^{m+1}} \ln\Gamma(z)$
$\displaystyle(4)\ \ \int\ln\Gamma(z)\ dz=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +z\log\Gamma(z) -\log \text{G}(1+z)+C$,
$\qquad$where $\text{G}(\cdot)$ is the Barnes G-function and $C$ is a constant of integration. Also for $a,b>0$
$$
\int\ln\Gamma\left(\frac{x+a}{b}\right)\ dx=b\ \psi^{(-2)}\left(\frac{x+a}{b}\right)+C.
$$
$\displaystyle(5)\ \ $As $\alpha\to\infty$, both of sides tend to $0$, hence the constant of integration is $0$.
A: $$I=\frac{\ln2}6+\frac{\ln\pi}2-6\ln A+\frac2\pi G,$$
where $A$ is the Glaisher–Kinkelin constant and $G$ is the Catalan constant.
A: Let's give a (sketch of the) proof of the following closed form evaluation.

$$
I=\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx=\frac{\ln2}6+\frac{\ln\pi}2-6\ln A+\frac2\pi G,
$$

where $A$ is the Glaisher–Kinkelin constant and $G$ is the Catalan constant.
Observe that, by the change of variable $\displaystyle u=\frac 1x$, we have
$$
I=\int_0^1\frac{u-1-u\ln u}{\left(1+u^2\right) \ln^2 u} \mathrm du=\int_0^1\frac{(1-u^2)(u-1-u\ln u)}{\left(1-u^4\right) \ln^2 u} \mathrm du.
$$
We set
$$
I(s):=\int_0^1u^s\frac{(1-u^2)(u-1-u\ln u)}{\left(1-u^4\right) \ln^2 u} \mathrm du, \quad s\geq0.
$$
We are allowed to differentiate $I(s)$  twice to obtain
$$
I''(s)=\int_0^1u^s\frac{(1-u^2)(u-1-u\ln u)}{\left(1-u^4\right)} \mathrm du. 
$$
Using the standard expansion $\displaystyle \frac{1}{1-u^4}=\sum_{k=0}^{\infty}u^{4k}, \, |u|<1,$ and performing the termwise integration, 
$$
\begin{align}
I''(s)=\sum_{k=0}^{\infty}\int_0^1u^{s+4k}(1-u^2)(u-1-u\ln u) \mathrm du
\end{align}
$$ 
gives
$$
\begin{align}
I''(s)=\sum_{k=0}^{\infty}\left(\frac{1}{(4k+2+s)^2}-\frac{1}{(4k+4+s)^2}+\frac{1}{(4k+2+s)}-\frac{1}{(4k+1+s)}\right).
\end{align}
$$ 
Now, recall the following series representation of the digamma function $\displaystyle \psi : = \Gamma'/\Gamma$,
$$
\psi(u+1) = -\gamma + \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{u+k}  
\right), \quad u >-1, 
$$ where $\gamma$ is the Euler-Mascheroni constant and, by differentiation, giving$$
\psi'(u+1) = \sum_{k=1}^{\infty} \frac{1}{(u+k)^2}, \quad u>-1.
$$
Hence
$$
I''(s)=\frac{1}{16}\psi'\left(\frac{s+2}{4}\right)-\frac{1}{16}\psi'\left(\frac{s+4}{4}\right)+\frac{1}{4}\psi\left(\frac{s+4}{4}\right)+\frac{1}{4}\psi\left(\frac{s+1}{4}\right)-\frac{1}{4}\psi\left(\frac{s+3}{4}\right)-\frac{1}{4}\psi\left(\frac{s+2}{4}\right).
$$
We have, as $s \rightarrow +\infty$, $I(s) \rightarrow 0$ and $I'(s) \rightarrow 0$, leading to
$$
I(s)=\log \Gamma \left(\frac{s+2}{4}\right)-\log \Gamma \left(\frac{s+4}{4}\right) +  
 4 \left(\psi\left(-2, \frac{s+4}{4}\right) + \psi\left(-2, \frac{s+1}{4}\right) - 
    \psi\left(-2, \frac{s+2}{4}\right) - \psi\left(-2, \frac{s+3}{4}\right)\right),
$$
then, with the use of Wolfram Alpha, as $s \rightarrow 0$ we get
$$
I=I(0)=\frac{\ln2}6+\frac{\ln\pi}2-6\ln A+\frac2\pi G.
$$
