Monstrous logarithm integral

Similar solution as @Henri Kärpijoki 's, presented as follows:

Using differentiation under the integration sign , define $I(a,t)$ as follows:

$\begin{aligned} I(a,t) & = \int_0^\infty \frac {\ln(tx^2 +1)}{x^2(x^2+a)} dx \\ \frac {\partial I(a,t)}{\partial t} & = \int_0^\infty \frac 1{(x^2+a)(tx^2+1)} dx \\ & = \frac 1{1-at} \int_0^\infty \left(\frac 1{x^2+a} - \frac t{tx^2+1} \right) dx \\ & = \frac 1{1-at} \left[\frac {\tan^{-1} \left(\frac x{\sqrt a}\right)}{\sqrt a} - \sqrt t \tan^{-1} \left(\sqrt t x \right) \right]_0^\infty \\ & = \frac {\pi\left(1-\sqrt{at}\right)}{2\sqrt a(1-at)} = \frac \pi{2\sqrt a(1+\sqrt{at})} & \small \blue{\text{Since }1-at = (1-\sqrt{at})(1+\sqrt{at})} \\ \implies I(a,t) & = \int \frac \pi{2\sqrt a(1+\sqrt{at})} \ dt & \small \blue{\text{Let }u^2 = at \implies 2u\ du = a\ dt} \\ & = \frac \pi{a\sqrt a} \int \frac u{1+u}du = \frac \pi{a\sqrt a} \int \left(1-\frac 1{1+u}\right) du \\ & = \frac {\pi (u-\ln (1+u))}{a\sqrt a} + \blue C & \small \blue{\text{where }C \text{ is the constant of integration.}} \\ & = \frac {\pi (\sqrt{at}-\ln (1+\sqrt{at}))}{a\sqrt a} & \small \blue{\text{Since }I(a,0) = 0 \implies C = 0} \\ \implies I(a) & = \frac {\pi (\sqrt a-\ln (1+\sqrt a))}{a\sqrt a} & \small \blue{\text{Putting }t=1} \end{aligned}$

Therefore,

$\begin{aligned} I(1) + I(2) & = \pi(1-\ln 2) + \frac \pi{2\sqrt 2}\left(\sqrt 2 - \ln \left(\sqrt 2 + 1\right) \right) \\ & = \frac \pi 4 \left(6-\sqrt 2 \ln \left(\sqrt 2 + 1\right) - 4\ln 2 \right) \end{aligned}$

Then we have $x_1 + x_2 + x_3 + x_4 = 2 + 1 + 3 + 4 = \boxed{10}$ .

$\ I\left(a\right)=\int_0^{\infty}\frac{\ln\left(x^2+1\right)}{x^2\left(x^2+a\right)}dx$

Let's define a new integral:

$I\left(a{,}t\right)=\int_0^{\infty}\frac{\ln\left(tx^2+1\right)}{x^2\left(x^2+a\right)}dx$

Let's use differentiation under the integral sign

$\Rightarrow\frac{d}{dt}I\left(a{,}t\right)=\frac{d}{dt}\int_0^{\infty}\frac{\ln\left(tx^2+1\right)}{x^2\left(x^2+a\right)}dx=\int_0^{\infty}\frac{\partial}{\partial t}\frac{\ln\left(tx^2+1\right)}{x^2\left(x^2+a\right)}dx=\int_0^{\infty}\frac{1}{tx^2+1}\cdot\frac{1}{x^2+a}\ dx$

Then we have to use partial fraction decomposition and get integral in the form:

$\frac{d}{dt}I\left(a{,}t\right)=\int_0^{\infty}\frac{t}{at-1}\cdot\frac{1}{\left(x\sqrt{t}\right)^2+1}-\frac{1}{at-1}\cdot\frac{1}{x^2+a}\ dx$

Let's use the integration formula $\int_{ }^{ }\frac{1}{\left(ax\right)^2+1}dx=\frac{1}{a}\arctan\left(ax\right)+C$

Then we are going to end up with

$\frac{d}{dt}I\left(a{,}t\right)=\frac{t}{at-1}\cdot\frac{\pi}{2\sqrt{t}}-\frac{1}{at-1}\cdot\frac{\pi}{2\sqrt{a}}$

We notice that $I\left(a{,}0\right)=0\wedge I\left(a{,}1\right)=I\left(a\right)$

$\int_0^1\frac{d}{dt}I\left(a{,}t\right)\ dt=I\left(a{,}1\right)=I\left(a\right)=\frac{\pi}{2}\int_0^1\frac{t}{at-1}\cdot\frac{1}{\sqrt{t}}-\frac{1}{at-1}\cdot\frac{1}{\sqrt{a}}\ dt$

Solving this trivial integral (you can use u sub to solve both integrals) we are going to end up with

$I\left(a\right)=\pi\cdot a^{-\frac{3}{2}}\left(\sqrt{a}-\ln\left(\sqrt{a}+1\right)\right)$

Then we calculate the sum of the first two values:

$I\left(1\right)=\pi\cdot1^{-\frac{3}{2}}\cdot\left(\sqrt{1}-\ln\left(\sqrt{1}+1\right)\right)=\pi\cdot\left(1-\ln\left(2\right)\right)$

$I\left(2\right)=\pi\cdot2^{-\frac{3}{2}}\cdot\left(\sqrt{2}-\ln\left(\sqrt{2}+1\right)\right)=\frac{\pi}{2\sqrt{2}}\left(\sqrt{2}-\ln\left(\sqrt{2}+1\right)\right)$

$I\left(1\right)+I\left(2\right)=\pi\cdot\left(1-\ln\left(2\right)\right)+\frac{\pi}{2\sqrt{2}}\left(\sqrt{2}-\ln\left(\sqrt{2}+1\right)\right)$

$=\frac{\pi}{4}\left(6-4\ln\left(2\right)-\sqrt{2}\ln\left(\sqrt{2}+1\right)\right)=-\frac{\pi}{4}\left(\sqrt{2}\ln\left(\sqrt{2}+1\right)+4\ln\left(2\right)-6\right)$

$=-\left(\sqrt{2}\cdot\ln\left(\sqrt{2}+1\right)+2^2\cdot\ln\left(2\right)-3\cdot2\right)\cdot\frac{\pi}{4}$

Finally, we get

$\Rightarrow x_1=2\wedge x_2=1\wedge x_3=3\wedge x_4=4\Rightarrow\sum_{1\le i\le4}^{ }x_i=10$