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.
$$ \begin{align}  G&= \int_{0}^{1} \int_{0}^{1} \frac{1}{2-x^{2}-y^{2}} \, dx \, dy \\ &= \int_{0}^{1} \frac{1}{2-y^{2}} \int_{0}^{1} \frac{1}{1-\frac{x^{2}}{2-y^{2}}} \, dx \, dy \\ &= \int_{0}^{1} \frac{1}{\sqrt{2-y^{2}}} \,  \text{artanh} \left(\frac{1}{\sqrt{2-y^{2}}} \right) \, dy \\ &= \int_{0}^{\pi/4} \text{artanh} \left(\frac{1}{\sqrt{2} \cos \theta} \right) \, d \theta \\ &= \frac{1}{2} \int_{0}^{\pi/4} \ln \left(\frac{\sqrt{2} \cos \theta +1}{\sqrt{2} \cos \theta -1} \right) \, d \theta \\ &= \frac{1}{2} \int_{0}^{\pi/4} \ln \left(\frac{(\sqrt{2} \cos \theta+1)^{2}}{2  \cos^{2} \theta -1} \right) \, d \theta \\ &=\int_{0}^{\pi/4} \ln (\sqrt{2} \cos \theta +1) \, d \theta - \frac{1}{2} \int_{0}^{\pi/4} \ln(\cos 2 \theta) \, d \theta \\ &= \int_{0}^{\pi/4} \ln\left(\sqrt{2} \cos \left(\frac{\pi}{4}  - \phi\right)+1\right) \, d \phi - \frac{1}{4} \int_{0}^{\pi/2} \ln( \cos \tau) \, d \tau \\ &= \int_{0}^{\pi/4} \ln \left(\sin(\phi) + \cos(\phi)+1\right) \, d \phi - \frac{1}{4} \left(- \frac{\pi}{2} \,  \ln 2 \right) \\ &= \int_{0}^{\pi/4} \ln (\sin \phi) \, d \phi + \int_{0}^{\pi/4} \ln \left(1+ \frac{1+ \cos \phi}{\sin \phi} \right) \, d \phi + \frac{\pi}{8} \,  \ln 2 \\ &= - \frac{G}{2} - \frac{\pi}{4} \,  \ln 2 + \int_{0}^{\pi/4} \ln \left(1+ \cot \frac{\phi}{2} \right) \, d \phi + \frac{\pi}{8} \, \ln 2 \\ &=  - \frac{G}{2} - \frac{\pi}{8} \, \ln 2 + 2 \int_{0}^{\pi/8} \ln (1 + \cot u) \, du \end{align}$$

Using the following trigonometric identitiy:
$$(1+\cot x)^2 =\frac{ 2 \tan\left(\frac{\pi}{4}+x\right)}{\tan(2 x) \, \tan(x)},$$
We have 
\begin{align}2 \int_0^{\pi/8} \ln(1+ \cot x) \, dx=
 & \ln2 \int_0^{\pi/8} dx-\int_0^{\pi/8} \ln \tan 2 x \, dx
  \\ - 
 & \int_0^{\pi/8} \ln \tan x \, dx+\int_0^{\pi/8} \ln \tan\left(\frac{\pi}{4}+x\right)\, dx \tag{1}
  \\ =
  & \frac{\pi}{8} \ln 2-\frac12 \int_0^{\pi/4} \ln \tan x \, dx
 \\ -
  & \int_0^{\pi/8} \ln \tan x \, dx-\int_{\pi/8}^{\pi/4} \ln \tan x \, dx \tag{2}
  \\ =
  & \frac{\pi}{8} \ln 2-\frac32  \int_0^{\pi/4} \ln \tan x \, dx 
  \\=
  &  \frac{\pi}{8} \ln 2 + \frac32 G.
 \end{align}
Where, to arrive at $(2)$ from $(1)$ we have substituted $x \mapsto x/2$ in the second integral, and in the last integral we
substituted $x \mapsto \pi/4-x$, in tandem with the fact that $\tan(\pi/2-x) = 1/\tan(x)$.
Also, the integral $\int_0^{\pi/4} \ln \tan x \, dx = - G$ is well known and can be easily reproduced.
