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Relevant wiki: Catalan's constant

$\begin{aligned} I & = \int_1^\infty \frac {\ln (1+x^2)}{1+x^2} dx & \small \color{#3D99F6} \text{Let }x = \tan \theta \implies dx = \sec^2 \theta \ d\theta \\ & = \int_\frac \pi 4^\frac \pi 2 \ln (\sec^2 \theta) \ d\theta \\ & = - 2 \int_\frac \pi 4^\frac \pi 2 \ln (\cos \theta) \ d\theta \\ & = - 2 \left({\color{#3D99F6}\int_0^\frac \pi 2 \ln (\cos \theta) \ d\theta} - \int_0^\frac \pi 4 \ln (\cos \theta) \ d\theta \right) & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = - 2{\color{#3D99F6}\int_0^\frac \pi 2 \ln (\sin \theta) \ d\theta} + 2 \int_0^\frac \pi 4 \ln (\cos \theta) \ d\theta & \small \color{#3D99F6} \text{Let } \phi = \frac \theta 2 \implies d \phi = \frac {d\theta}2 \\ & = - 4{\color{#3D99F6}\int_0^\frac \pi 4 \ln (2\sin \phi \cos \phi) \ d\phi} + 2 \int_0^\frac \pi 4 \ln (\cos \theta) \ d\theta & \small \color{#3D99F6} \text{Replace } \phi \text{ with } \theta \\ & = {\color{#3D99F6}- 4 \int_0^\frac \pi 4 \ln (2\sin \theta) \ d\theta} - 2 \int_0^\frac \pi 4 \ln (\cos \theta) \ d\theta & \small \color{#3D99F6} \text{Catalan's constant }G = - 2\int_0^\frac \pi 4 \ln(2\sin x)\ dx \\ & = {\color{#3D99F6}2G} - {\color{#D61F06} 2 \int_0^\frac \pi 4 \ln (2\cos \theta) \ d\theta} + 2 \int_0^\frac \pi 4 \ln 2 \ d\theta & \small \color{#D61F06} \text{Catalan's constant }G = 2\int_0^\frac \pi 4 \ln(2\cos x)\ dx \\ & = {\color{#3D99F6}G} + \frac \pi 2 \ln 2 & \small \color{#3D99F6} G \approx 0.91597 \\ & \approx \boxed{2.005} \end{aligned}$