TITLE: What is value of this integral? $\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^{2})}dx$
QUESTION [10 upvotes]: What is value of this integral $$I=\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^2)}dx$$

My work :
\begin{align*}I&=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)(10+10x^2)}{(9+x^2)(1+9x^2)}dx\\
&=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)}{1+9x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)}{9+x^2}dx\\
&=j_{1}+j_{2}+j_{3}\\
\end{align*}
$$j_{1}=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx=\sum_{n=0}^{\infty}\frac{(-1)^n(2^{2(n+1)})}{n+1}\int_{0}^{\infty}\frac{x^{2(n+1)}}{9+x^2}dx$$ 
$$j_{2}=\log(4)\int_{0}^{\infty}\frac{1}{1+(3x)^2}dx+\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(2^{2(n+1)})}\int_{0}^{\infty}\frac{x^{2(n+1)}}{1+9x^2}dx=\frac{\log(4)\pi}{6}+\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)2^{2(n+1)}}\int_{0}^{\infty}\frac{x^{2(n+1)}}{1+9x^2}dx$$
$$j_{3}=\int_{0}^{\infty}\frac{\log(4+x^2)}{9+x^2}dx=\frac{\log(4)\pi}{54}+\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)2^{2(n+1)}}\int_{0}^{\infty}\frac{x^{2(n+1)}}{9+x^2}dx$$
Wait for a review to find solutions to this

REPLY [4 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{a,b,c \in \mathbb{R}_{\ >\ 0}}$:
\begin{align}
\mrm{f}\pars{a,b,c} & \equiv
\int_{0}^{\infty}{\ln\pars{a^{2} + b^{2}x^{2}} \over c^{2} + x^{2}}\,\dd x =
\Re\int_{-\infty}^{\infty}{\ln\pars{a + bx\ic} \over
\pars{x + c\ic}\pars{x - c\ic}}\,\dd x
\end{align}
With the change
$\ds{\pars{x = -\,{s - a \over b}\,\ic \implies s = a + bx\ic}}$:
\begin{align}
\mrm{f}\pars{a,b,c} & =
-\,\Im\int_{a - \infty\ic}^{a + \infty\ic}{b\ln\pars{s} \over
\bracks{s - \pars{a - bc}}\bracks{s - \pars{a + bc}}}\,\dd s
\\[5mm] & =
-\,\Im\bracks{-2\pi\ic\,{b\ln\pars{a + bc} \over 2bc}} =
\bbx{\pi\,{\ln\pars{a + bc} \over c}} \\ & 
\end{align}

$$
\left\{\begin{array}{rcccccl}
\ds{j_{1}} & \ds{=} &
\ds{\int_{0}^{\infty}{\ln\pars{1 + 4x^{2}} \over 9 + x^{2}}\,\dd x}
& \ds{=} & \ds{\mrm{f}\pars{1,2,3}} & \ds{=} & \ds{{\pi \over 3}\ln\pars{7}}
\\[2mm]
\ds{j_{2}} & \ds{=} &
\ds{\int_{0}^{\infty}{\ln\pars{4 + x^{2}} \over 1 + 9x^{2}}\,\dd x}
& \ds{=} & \ds{{1 \over 9}\,\mrm{f}\pars{2,1,{1 \over 3}}} & \ds{=} & \ds{{\pi \over 3}\,\ln\pars{7 \over 3}}
\\[2mm]
\ds{j_{3}} & \ds{=} &
\ds{\int_{0}^{\infty}{\ln\pars{4 + x^{2}} \over 9 + x^{2}}\,\dd x}
& \ds{=} & \ds{\mrm{f}\pars{2,1,3}} & \ds{=} 
& \ds{{\pi \over 3}\,\ln\pars{5}}
\end{array}\right.
$$
\begin{align}
\mbox{} &
\\
j_{1} + j_{2} + j{3} & =
\bbx{{1 \over 3}\,\pi\ln\pars{245 \over 3}} \approx 4.6104 \\ &
\end{align}