Prove that $\int_{0}^{\infty}{\operatorname{arccot}^2(\sqrt{1+x})\over 1+x}\mathrm dx=\pi C-{7\over 4}\zeta(3)$ 
$$I=\int_{0}^{\infty}{\operatorname{arccot}^2(\sqrt{1+x})\over 1+x}\mathrm dx$$
$$J=4\int_{0}^{\infty}{\operatorname{arccot}^2[(1+x)^2
]\over 1+x}\mathrm dx$$

Q1: Show that $I=J\color{red}?$
Q2: How can we prove that $$I=\pi C-{7\over 4}\zeta(3)\color{red}?$$
Where $C$ is Catalan's constant

$u=\sqrt{1+x}\implies 2udu=dx$
$$I=2\int_{1}^{\infty}{\operatorname{arccot}^2(u)\over u}\mathrm du$$
$v=(1+x)^2\implies dv=2(1+x)dx$
$$J=2\int_{1}^{\infty}{\operatorname{arccot}^2(v)\over v}\mathrm dv$$
 A: You have already showed that $I=J$: they are associated with the same integral, i.e.
$$ I=2\int_{0}^{1}\frac{\arctan^2(u)}{u}\,du = 4 \int_{0}^{\pi/4}\frac{t^2}{\sin(2t)}\,dt=\frac{1}{2}\int_{0}^{\pi/2}\frac{z^2}{\sin z}\,dz\tag{1}$$
The claim now follows from the Eisenstein series for $\frac{1}{\sin z}$:
$$ \frac{1}{\sin z} = \frac{1}{z}+\sum_{n\geq 1}(-1)^n\left(\frac{1}{z-n\pi}+\frac{1}{z+n\pi}\right) \tag{2}$$
by multiplying both sides by $z^2$, performing integration over $(0,\pi/2)$ and rearranging.
An efficient alternative is to use a step of integration by parts to get $I=\int_{0}^{\pi/2}z\log\tan\frac{z}{2}\,dz$, then exploit the Fourier series of $\log\sin$ and $\log\cos$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\int_{0}^{\infty}{\mrm{arccot}^{2}\pars{\root{1 + x}} \over 1 + x}\,\dd x
\,\,\,\stackrel{2\,\mrm{arccot}\pars{\!\!\root{1 + x}\!\!}\ \mapsto\ x}{=}\,\,\,
{1 \over 2}\int_{0}^{\pi/2}{x^{2} \over \sin\pars{x}}\,\dd x
\\[5mm] = &\
\left.{1 \over 2}\,\Re\int_{x = 0}^{x = \pi/2}
{-\ln^{2}\pars{z} \over \pars{1 - z^{2}}\ic/\pars{2z}}\,{\dd z \over \ic z}
\right\vert_{\ z\ =\ \exp\pars{\ic x}} =
\left.\Re\int_{x = 0}^{x = \pi/2}
{\ln^{2}\pars{z} \over 1 - z^{2}}\,\dd z\,
\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] = &\
-\,\Re\int_{1}^{0}
{\bracks{\ln\pars{y} + \pi\ic/2}^{\,2} \over 1 + y^{2}}\,\ic\,\dd y -
\Re\int_{0}^{1}
{\ln^{2}\pars{x} \over 1 - x^{2}}\,\dd x
\\[5mm] = &\
-\pi\int_{0}^{1}{\ln\pars{x} \over 1 + x^{2}}\,\dd x -
\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x^{2}}\,\dd x
\\[5mm] = &\
-\pi\
\underbrace{\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{2}}\,\dd x}
_{\ds{\substack{\ds{=\ {\large 0}}}\ \mbox{because}\\ \mbox{it}\ changes\ its\ sign\\ \mbox{under}\ x\ \to\ 1/x}}\ +\
\pi\ \underbrace{\int_{1}^{\infty}{\ln\pars{x} \over 1 + x^{2}}\,\dd x}
_{\ds{\substack{{\large =\ G}\\[1mm]\color{#f00}{\Large\S}:\  Catalan\ Constant}}}\ -\
\underbrace{\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x^{2}}\,\dd x}
_{\ds{{7 \over 4}\,\zeta\pars{3}}}\
\\[5mm] = &\ 
\bbx{\pi\,G - {7 \over 4}\,\zeta\pars{3}}
\end{align}

$\ds{\color{#f00}{\large\S}}$: See
  Catalan Constant Page.

The last integral is evaluated as follows:
\begin{align}
\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x^{2}}\,\dd x & =
{1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x}\,\dd x +
{1 \over 2}\int_{-1}^{0}{\ln^{2}\pars{-x} \over 1 - x}\,\dd x
\\[5mm] & \stackrel{\mbox{IBP}}{=}\,\,\,
-\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln\pars{x}\,\dd x -\int_{-1}^{0}\mrm{Li}_{2}'\pars{x}\ln\pars{-x}\,\dd x
\\[5mm] & \stackrel{\mbox{IBP}}{=}\,\,\,
\int_{0}^{1}\mrm{Li}_{3}'\pars{x}\,\dd x +
\int_{-1}^{0}\mrm{Li}_{3}'\pars{x}\,\dd x =
\mrm{Li}_{3}\pars{1} - \mrm{Li}_{3}\pars{-1}
\\[5mm] & =
\zeta\pars{3} - \bracks{-\,{3 \over 4}\,\zeta\pars{3}} =
{7 \over 4}\,\zeta\pars{3}
\end{align}
