What is value of this integral? $\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^{2})}dx$ 
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
 A: First, the integral:
$$
I := \int_0^{\infty} \frac{\ln\left(1+4\,x^2\right)\left(1+9\,x^2\right)\left(9+\,x^2\right)+\left(9+x^2\right)\ln\left(4+x^2\right)\left(10+10\,x^2\right)}{\left(9+x^2\right)^2\left(1+9\,x^2\right)}\,\text{d}x
$$
thanks to the linearity of the integrals it can be written as the algebraic sum of two integrals:
$$
I = 
\int_0^{\infty} \frac{\ln\left(1+4\,x^2\right)}{9+x^2}\,\text{d}x + 
\int_0^{\infty} \frac{10\left(1+x^2\right)\ln\left(4 + x^2\right)}{\left(9+\,x^2\right)\left(1+9\,x^2\right)}\,\text{d}x\,.
$$
At this point, to calculate the first, it's sufficient to consider the function $J : [0,\,+\infty) \to \mathbb{R}$ of the law:
$$
J(a) := \int_0^{\infty} \frac{\ln\left(1+a\,x^2\right)}{9+x^2}\,\text{d}x
$$
whose first derivative is equal to:
$$
J'(a) = \int_0^{\infty} \frac{x^2}{\left(9+x^2\right)\left(1+a\,x^2\right)}\,\text{d}x = \frac{\pi}{2\left(\sqrt{a}+3\,a\right)}
$$
and therefore, going back, we get:
$$
J(a) = \int \frac{\pi}{2\left(\sqrt{a}+3\,a\right)}\,\text{d}a = \frac{\pi}{3}\,\ln\left(1 + 3\sqrt{a}\right) + c
$$
where equaling the two expressions of $J(0)$ must be $c = 0$, from which:
$$
\int_0^{\infty} \frac{\ln\left(1+a\,x^2\right)}{9+x^2}\,\text{d}x = \frac{\pi}{3}\,\ln\left(1 + 3\sqrt{a}\right).
$$
Similarly, to calculate the second, it's sufficient to consider the function $K : [0,\,+\infty) \to \mathbb{R}$ of the law:
$$
K(b) := \int_0^{\infty} \frac{10\left(1+x^2\right)\ln\left(4 + b\,x^2\right)}{\left(9+\,x^2\right)\left(1+9\,x^2\right)}\,\text{d}x
$$
whose first derivative is equal to:
$$
K'(b) = \int_0^{\infty} \frac{10\,x^2\left(1+x^2\right)}{\left(9+x^2\right)\left(1+9\,x^2\right)\left(4+b\,x^2\right)}\,\text{d}x = \frac{\left(3+\frac{10}{\sqrt{b}}\right)\pi}{3\left(12+20\sqrt{b}+3\,b\right)}
$$
and therefore, going back, we get:
$$
K(b) = \int \frac{\left(3+\frac{10}{\sqrt{b}}\right)\pi}{3\left(12+20\sqrt{b}+3\,b\right)}\,\text{d}b = \frac{\pi}{3}\,\ln\left(12+20\sqrt{b}+3\,b\right) + c
$$
where equaling the two expressions of $K(0)$ must be $c = \frac{2\,\pi}{3}\,\ln(2) - \frac{\pi}{3}\,\ln(12)$, from which:
$$
\int_0^{\infty} \frac{10\left(1+x^2\right)\ln\left(4 + b\,x^2\right)}{\left(9+\,x^2\right)\left(1+9\,x^2\right)}\,\text{d}x = \frac{\pi}{3}\,\ln\left(12+20\sqrt{b}+3\,b\right) + \frac{2\,\pi}{3}\,\ln(2) - \frac{\pi}{3}\,\ln(12)\,.
$$
In conclusion, setting $a = 4$ and $b = 1$, we get:
$$
I = \frac{\pi}{3}\,\ln\left(1 + 3\sqrt{4}\right) + \frac{\pi}{3}\,\ln\left(12+20\sqrt{1}+3\cdot 1\right) + \frac{2\,\pi}{3}\,\ln(2) - \frac{\pi}{3}\,\ln(12) = \frac{\pi}{3}\,\ln\left(\frac{245}{3}\right),
$$
which is what is desired.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}}}
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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}
A: Consider using differentiation under the integral sign.  Parameterize the integral as the following:
$$I(a)=\int_0^{\infty} \frac{\ln{\left(1+ax^2\right)}}{9+x^2} + \frac{\ln{\left(a+x^2\right)}}{1+9x^2}+ \frac{\ln{\left(a+x^2\right)}}{9+x^2} \; \mathrm{d}x$$
The integral in question is $I(4)$.  First, differentiate $I(a)$ with respect to $a$: \begin{align*}
I'(a)&=\int_0^{\infty} \frac{x^2}{(9+x^2)(1+ax^2)}+\frac{1}{(1+9x^2)(a+x^2)}+\frac{1}{(9+x^2)(a+x^2)} \; \mathrm{d}x \\
&=\int_0^{\infty} -\frac{10 (a - 1)}{(a - 9) (9 a - 1) (a + x^2)} + \frac{2 (9 a - 41)}{(a - 9) (9 a - 1) (x^2 + 9)} + \frac{9}{(9 a - 1) (9 x^2 + 1)} + \frac{1}{(1 - 9 a) (a x^2 + 1)} \; \mathrm{d}x \\
&=\frac{\pi}{6} \left(\frac{\frac{19}{\sqrt{a}}+9}{3a+10\sqrt{a}+3}\right)\\
\end{align*}
Now, $I'(a)$ with respect to $a$ from $0$ to $4$:
\begin{align*}
I(4)&=\frac{\pi}{6} \int_0^4 \frac{\frac{19}{\sqrt{a}}+9}{3a+10\sqrt{a}+3} \; \mathrm{d}a \\
&=\frac{\pi}{6} \int_0^4 \frac{\frac{1}{\sqrt{a}}}{\sqrt{a}+3} + \frac{\frac{6}{\sqrt{a}}}{3\sqrt{a}+1} \; \mathrm{d} a \\
&=\frac{\pi}{6} \left(2 \int_0^4 \frac{\mathrm{d}\left(\sqrt{a}+3\right)}{\sqrt{a}+3} +  4 \int_0^4 \frac{\mathrm{d}\left(3\sqrt{a}+1\right)}{3\sqrt{a}+1}\right) \\
&=\frac{\pi}{3} \left(\ln{\left(\sqrt{a}+3\right)}+2\ln{\left(3\sqrt{a}+1\right)}\right) \bigg \rvert_0^4 \\
I(4) &= \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)} \mathrm{d}x =\boxed{\frac{\pi \ln{\left(\frac{245}{3}\right)}}{3}}\\
\end{align*}
