Ugly Integral

Relevant wiki: Integration Tricks

$\begin{aligned} I(t) & = \int_0^\infty \frac {\ln \left(1+\frac {t^2}{x^2}\right)}{1+x^2} dx & \small \color{#3D99F6} \text{Using }\int_0^\infty f(x) \ dx = \int_0^\infty \frac {f\left( \frac 1x \right)}{x^2} dx \\ & = \int_0^\infty \frac {\ln \left(1+t^2x^2\right)}{1+x^2} dx \\ \frac {\partial I(t)}{\partial t} & = \int_0^\infty \frac {2tx^2}{(1+x^2)(1+t^2x^2)} dx \\ & = \frac {2t}{t^2-1} \int_0^\infty \left(\frac 1{x^2+1} - \frac 1{t^2x^2+1}\right) dx \\ & = \frac {2t}{t^2-1} \left[ \tan^{-1} x - \frac {\tan^{-1} (tx)}t \right]_0^\infty \\ & = \frac \pi{t+1} \\ \implies I(t) & = \int \frac \pi{t+1} dt \\ & = \pi \ln (t+1) + C & \small \color{#3D99F6} \text{where } C \text{ is the constant of integration.} \\ I(0) & = \pi \ln(0+1) + C & \small \color{#3D99F6} \text{Note that } I(0) = 0 \implies C = 0 \\ \implies I(t) & = \pi \ln(t+1) = \pi \ln 2019 \\ \implies t & = \boxed{2018} \end{aligned}$

If we define $f(t) \; = \; \int_0^\infty \frac{\ln\big(1 + \frac{t^2}{x^2}\big)}{1+x^2}\,dx \hspace{2cm} t \ge 0$ then, for $t > 0$ , $\begin{aligned} f'(t) & = \; \int_0^\infty \frac{1}{1+x^2} \frac{1}{1 + \frac{t^2}{x^2}} \frac{2t}{x^2}\,dt \; = \; 2t\int_0^\infty\frac{dx}{(x^2+1)(x^2+t^2)} \\ & = \; \frac{2t}{t^2-1}\int_0^\infty \left(\frac{1}{x^2+1} - \frac{1}{x^2 + t^2}\right)\,dx \; = \; \frac{2t}{t^2-1}\Big[\tan^{-1}x - \tfrac{1}{t}\tan^{-1}\tfrac{x}{t}\big]_0^\infty \\ & = \; \frac{2t}{t^2-1} \big(1 - t^{-1}\big) \tfrac{\pi}{2} \; = \; \tfrac{\pi}{t+1} \end{aligned}$ Since $f(0) = 0$ we deduce that $f(t) = \pi\ln(t+1)$ . Thus the answer is $\boxed{2018}$ .