A different way to prove that $\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx = \frac{3 G}{4} + \frac{\pi}{16} \, \ln 2$ From the fact that $$\int_{0}^{1} \int_{0}^{1} \frac{1}{2-x^{2}-y^{2}} \, dx \, dy $$ is an integral representation of Catalan's constant ($G$), I was able to deduce that $$\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx = \int_{0}^{\pi/8} \ln(1+ \cot u) \, du = \frac{3G}{4} + \frac{\pi}{16} \, \ln 2\tag{1}.$$

What is another way to prove $(1)$ that preferably doesn't involve the dilogarithm function?


EDIT:
In response to Dr. MV's comment, the following is how I deduced $(1)$ from that integral representation of Catalan's constant.
$$ \begin{align}  G&= \int_{0}^{1} \int_{0}^{1} \frac{1}{2-x^{2}-y^{2}} \, dx \, dy \\ &= \int_{0}^{1} \frac{1}{2-y^{2}} \int_{0}^{1} \frac{1}{1-\frac{x^{2}}{2-y^{2}}} \, dx \, dy \\ &= \int_{0}^{1} \frac{1}{\sqrt{2-y^{2}}} \,  \text{artanh} \left(\frac{1}{\sqrt{2-y^{2}}} \right) \, dy \\ &= \int_{0}^{\pi/4} \text{artanh} \left(\frac{1}{\sqrt{2} \cos \theta} \right) \, d \theta \\ &= \frac{1}{2} \int_{0}^{\pi/4} \ln \left(\frac{\sqrt{2} \cos \theta +1}{\sqrt{2} \cos \theta -1} \right) \, d \theta \\ &= \frac{1}{2} \int_{0}^{\pi/4} \ln \left(\frac{(\sqrt{2} \cos \theta+1)^{2}}{2  \cos^{2} \theta -1} \right) \, d \theta \\ &=\int_{0}^{\pi/4} \ln (\sqrt{2} \cos \theta +1) \, d \theta - \frac{1}{2} \int_{0}^{\pi/4} \ln(\cos 2 \theta) \, d \theta \\ &= \int_{0}^{\pi/4} \ln\left(\sqrt{2} \cos \left(\frac{\pi}{4}  - \phi\right)+1\right) \, d \phi - \frac{1}{4} \int_{0}^{\pi/2} \ln( \cos \tau) \, d \tau \\ &= \int_{0}^{\pi/4} \ln \left(\sin(\phi) + \cos(\phi)+1\right) \, d \phi - \frac{1}{4} \left(- \frac{\pi}{2} \,  \ln 2 \right) \\ &= \int_{0}^{\pi/4} \ln (\sin \phi) \, d \phi + \int_{0}^{\pi/4} \ln \left(1+ \frac{1+ \cos \phi}{\sin \phi} \right) \, d \phi + \frac{\pi}{8} \,  \ln 2 \\ &= - \frac{G}{2} - \frac{\pi}{4} \,  \ln 2 + \int_{0}^{\pi/4} \ln \left(1+ \cot \frac{\phi}{2} \right) \, d \phi + \frac{\pi}{8} \, \ln 2 \\ &=  - \frac{G}{2} - \frac{\pi}{8} \, \ln 2 + 2 \int_{0}^{\pi/8} \ln (1 + \cot u) \, du \end{align}$$

Perform the change of variable $y=\arctan x$,
$\displaystyle I=\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx=\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln(1+\tan x)dx=\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}} \ln\left(\sin x+\cos x\right)dx-\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln\left(\cos x\right)dx$
$\begin{align}\displaystyle \int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}} \ln\left(\sin x+\cos x\right)dx&=\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln\left(\sqrt{2}\cos\left(\dfrac{\pi}{4}-x\right)\right)dx\\
&=\dfrac{\pi}{16}\ln 2+\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln\left(\cos\left(\dfrac{\pi}{4}-x\right)\right)dx\\
&=\dfrac{\pi}{16}\ln 2+\int_{\tfrac{\pi}{8}}^{\tfrac{\pi}{4}}\ln(\cos x)dx\\
&=\dfrac{\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\int_{0}^{\tfrac{\pi}{8}}\ln(\cos x)dx
\end{align}$
Perform the change of variable $y=\dfrac{\pi}{2}-x$,
$\displaystyle \int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln\left(\cos x\right)dx=\int_{0}^{\tfrac{\pi}{8}}\ln\left(\sin x\right)dx$
Therefore,
$\begin{align}
\displaystyle I&=\dfrac{\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\int_{0}^{\tfrac{\pi}{8}}\ln(\cos x)dx-\int_{0}^{\tfrac{\pi}{8}}\ln\left(\sin x\right)dx\\
&=\dfrac{\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\int_{0}^{\tfrac{\pi}{8}}\ln(\cos x\sin x)dx\\
&=\dfrac{\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\int_{0}^{\tfrac{\pi}{8}}\ln\left(\dfrac{\sin(2x)}{2}\right)dx\\
&=\dfrac{\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\dfrac{1}{2}\int_{0}^{\tfrac{\pi}{4}}\ln\left(\dfrac{\sin x}{2}\right)dx\\
&=\dfrac{3\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\dfrac{1}{2}\int_{0}^{\tfrac{\pi}{4}}\ln\left(\sin x\right)dx\\
&=\dfrac{3\pi}{16}\ln 2+\dfrac{1}{2}\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx+\dfrac{1}{2}\int_{0}^{\tfrac{\pi}{4}}\ln(\cot x)dx
\end{align}$
Since,
$\displaystyle \int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx=\dfrac{1}{2} G-\dfrac{1}{4}\pi\ln 2$
and,
$\displaystyle \int_{0}^{\tfrac{\pi}{4}}\ln(\cot x)dx=G$
then,
$\begin{align}
I&=\dfrac{3\pi}{16}\ln 2+\dfrac{1}{4} G-\dfrac{1}{8}\pi\ln 2+\dfrac{1}{2}G\\
&=\dfrac{\pi}{16}\ln 2+\dfrac{3}{4}G
\end{align}$
PS:
Proof of
$\displaystyle \int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx=\dfrac{1}{2} G-\dfrac{1}{4}\pi\ln 2$
$\begin{align}
\displaystyle \int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx+\int_0^{\tfrac{\pi}{4}}\ln(\cos x)dx&=\int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx+\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\ln(\sin x)dx\\
&=\int_0^{\tfrac{\pi}{2}}\ln(\sin x)dx\\
&=2\int_0^{\tfrac{\pi}{4}}\ln\left(\sin(2x)\right)dx\\
&=2\int_0^{\tfrac{\pi}{4}}\ln\left(2\sin x\cos x\right)dx\\
&=\dfrac{1}{2}\pi\ln 2+2\int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx+2\int_0^{\tfrac{\pi}{4}}\ln(\cos x)dx
\end{align}$
Therefore,
$\displaystyle \int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx+\int_0^{\tfrac{\pi}{4}}\ln(\cos x)dx=-\dfrac{1}{2}\pi\ln 2$
and,
$\displaystyle G=\int_0^{\tfrac{\pi}{4}}\ln(\cot x)dx=\int_0^{\tfrac{\pi}{4}}\ln(\cos x)dx-\int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx$
therefore,
$\displaystyle \int_0^{\tfrac{\pi}{4}}\ln(\cos x)dx=\dfrac{1}{2}G-\dfrac{1}{4}\pi\ln 2$
$\displaystyle \int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx=-\dfrac{1}{2}G-\dfrac{1}{4}\pi\ln 2$
