I will attempt to simplify the integral. It might take me a while and I might edit this answer a few times, so this is not its final form yet.
Let $D$ be the differential operator.
We know $\displaystyle\int \dfrac{dx}{\sqrt{x^2-1}}=\ln(2(x+\sqrt{x^2-1}))$.
So by using integration by parts we reduce the problem to solving $$\displaystyle\mathcal{A}=\int_1^\infty D\left[\operatorname{arccot}\left(1+\frac{2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)\right] \ln(2(x+\sqrt{x^2-1}))\, dx.$$
We know that $D[\mathrm{arccot}(x)] = \dfrac{-1}{1+x^2} $ hence by the chain rule for derivatives we get
$$\mathcal{A}=\int_1^\infty \dfrac{D\left[\left(\frac{-2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)\right] \ln(2(x+\sqrt{x^2-1}))}{1+\left(1+\dfrac{2\, \pi}{\mathrm{arcoth}(x) - \mathrm{arccsc}(x)}\right)^2}\,dx.$$
Because we know that $D[\mathrm{arccot}(x)] = \dfrac{-1}{1+x^2} $ and also that $ln(2z)=ln(2)+ln(z)$ we can use integration by parts again and arrive at
$$\mathcal{B}=\int_1^\infty \dfrac{D\left[\left(\frac{-2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)\right] \ln(x+\sqrt{x^2-1})}{1+\left(1+\dfrac{2\, \pi}{\mathrm{arcoth}(x) - \mathrm{arccsc}(x)}\right)^2}\,dx.$$
Now we can turn this into an infinite sum because we set $U=\dfrac{2\, \pi}{\mathrm{arcoth}(x) - \mathrm{arccsc}(x)}$ and use the Taylor expansion for $\frac{z}{1+(1+z)^2}$.
In other words we substitute $z=U$.
( Remember that $\int \Sigma = \Sigma \int$ )
Finally the core problem is reduced mainly to solving:
$$\displaystyle\mathcal{C_n}=\int_1^\infty \frac{\ln(x+\sqrt{x^2-1})}{(\mathrm{arcoth}(x)-\mathrm{arccsc}(x))^n}\, dx.$$
by induction we get the need to solve for $C_1$ and then are able to get the others.
At this point, I must admit that I have ignored convergeance issues.
Those issues can be solved by taking limits.
For instance $C_1$ does not actually converge by itself.
For all clarity the problem is not resolved.
In fact it might require a new bounty. Still thinking about it ...
(in progress)
Define the function $I(s)$ for $s > 0$ by
$$ I(s) = \int_{0}^{\infty} \frac{\operatorname{arccot}( \sqrt{x} - e^{s}\sqrt{x+1} )}{x+1} \, dx. $$
By observing that $I(\infty) = 0$, we have
$$ I(s) = -\int_{s}^{\infty} I'(t) \, dt. $$
Applying Leibniz's integral rule,
$$ I'(s) = \int_{0}^{\infty} \frac{e^{s}\sqrt{x+1}}{1 + (\sqrt{x} - e^{s}\sqrt{x+1} )^{2}} \, \frac{dx}{x+1}. $$
Now with the substitution $x = \tan^{2}\theta$,
\begin{align*}
I'(s)
&= \int_{0}^{\frac{\pi}{2}} \frac{\sin\theta}{\cosh s - \sin\theta}
= \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos\theta}{\cosh s - \cos\theta}.
\end{align*}
Then with the substitution $z = e^{i\theta}$, it follows that
\begin{align*}
I'(s)
&= \frac{i}{2} \int_{\Gamma} \frac{z^{2} + 1}{z^{2} - 2z \cosh s + 1} \frac{dz}{z},
\end{align*}
where the contour $\Gamma$ denotes the counter-clockwise semicircular arc joining from $-i$ to $i$. Note that we have $z^{2} - 2z \cosh s + 1 = 0$ if and only if $\cosh s = \cos \theta$ if and only if $z = e^{i\theta} = e^{\pm s}$. Thus deforming the contour to the straight line joining from $-i$ to $i$ with infinitesimal indent at the origin, we obtain
\begin{align*}
I'(s)
&= \frac{i}{2} \left\{ \operatorname{PV}\! \int_{-i}^{i} f(z) \, dz + 2\pi i \operatorname{Res} (f, e^{-s}) + \pi i \operatorname{Res} (f, 0) \right\},
\end{align*}
where $f$ denotes the integrand
$$ f(z) = \frac{z^{2} + 1}{z (z^{2} - 2z \cosh s + 1)}. $$
Proceeding the calculation,
\begin{align*}
I'(s)
&= \frac{i}{2} \left\{ i \operatorname{PV} \! \int_{-1}^{1} f(ix) \, dx + \pi i - 2 \pi i \coth s \right\} \\
&= - \frac{1}{2} \int_{0}^{1} \{ f(ix) + f(-ix) \} \, dx - \frac{\pi}{2} + \pi \coth s \\
&= \cosh s \int_{0}^{1} \frac{\frac{1}{2} (1 - x^{-2}) }{\{ \frac{1}{2}(x + x^{-1}) \}^{2} + \sinh^{2} s} \, dx - \frac{\pi}{2} + \pi \coth s.
\end{align*}
Now with the substitution $u = \frac{1}{2} \{ x + x^{-1} \}$, it follows that
\begin{align*}
I'(s)
&= - \cosh s \int_{1}^{\infty} \frac{du}{u^{2} + \sinh^{2} s} - \frac{\pi}{2} + \pi \coth s \\
&= - \arctan(\sinh s) \coth s - \frac{\pi}{2} + \pi \coth s.
\end{align*}
Finally, with the substitution $x = \sinh t$,
\begin{align*}
I(s)
= - \int_{s}^{\infty} I'(t) \, dt
&= \int_{s}^{\infty} \left\{ \arctan(\sinh t) \coth t + \frac{\pi}{2} - \pi \coth t \right\} \, dt \\
&= \int_{\sinh s}^{\infty} \left( \frac{\arctan x}{x} + \frac{\pi}{2\sqrt{x^{2} + 1}} - \frac{\pi}{x} \right) \, dx.
\end{align*}
Our problem corresponds to $s = \log 2$, or equivalently, $a := \sinh s = \frac{3}{4}$.
\begin{align*}
&\int_{a}^{\infty} \left( \frac{\arctan x}{x} + \frac{\pi}{2\sqrt{x^{2} + 1}} - \frac{\pi}{x} \right) \, dx \\
&= - \int_{a}^{\infty} \frac{\arctan (1/x)}{x} \, dx + \frac{\pi}{2} \int_{a}^{\infty} \left( \frac{1}{\sqrt{x^{2} + 1}} - \frac{1}{x} \right) \, dx \\
&= - \int_{0}^{1/a} \frac{\arctan x}{x} \, dx + \frac{\pi}{2} \lim_{R\to\infty} \left[ \log\left( \frac{x + \sqrt{x^{2} + 1}}{x} \right) \right]_{a}^{R} \\
&= - \operatorname{Ti}\left(\frac{1}{a}\right) - \frac{\pi}{2} \log\left( \frac{a + \sqrt{a^{2} + 1}}{2a} \right),
\end{align*}
where the function
$$ \operatorname{Ti}(x) = \frac{\operatorname{Li}_{2}(ix) - \operatorname{Li}_{2}(-ix)}{2i} $$
is the inverse tangent integral. Using the following simple identity
$$ \operatorname{Ti}(x) = \operatorname{Ti}\left(\frac{1}{x}\right) + \frac{\pi}{2}\log x, $$
it follows that
\begin{align*}
\int_{a}^{\infty} \left( \frac{\arctan x}{x} + \frac{\pi}{2\sqrt{x^{2} + 1}} - \frac{\pi}{x} \right) \, dx
= - \operatorname{Ti}(a) - \frac{\pi}{2} \log\left( \frac{a + \sqrt{a^{2} + 1}}{2a^{2}} \right).
\end{align*}
Therefore, plugging $a = \frac{3}{4}$ gives
$$ \int_{0}^{\infty} \frac{\operatorname{arccot}( \sqrt{x} - 2\sqrt{x+1} )}{x+1} \, dx
= \pi \log(3/4) - \operatorname{Ti}(3/4) $$
as Cleo pointed out without explanation. More generally, for $k > 1$
$$ \int_{0}^{\infty} \frac{\operatorname{arccot}( \sqrt{x} - k \sqrt{x+1} )}{x+1} \, dx
= \frac{\pi}{2} \log \left( \frac{(k^{2} - 1)^{2}}{2k^{3}} \right) - \operatorname{Ti}\left( \frac{k^{2} - 1}{2k} \right). $$
Best Answer
Splitting the integral
We can write \begin{align} I &\equiv \int \limits_0^\infty x \operatorname{arccot}(x) [1-x \operatorname{arccot}(x)][2+x \operatorname{arccot}(x)] \, \mathrm{d} x \\ &= \int \limits_0^\infty x \operatorname{arccot}(x) [1-x \operatorname{arccot}(x)] \, \mathrm{d} x + \int \limits_0^\infty x \operatorname{arccot}(x) [1-x^2 \operatorname{arccot}^2(x)] \, \mathrm{d} x \\ &= f_1 (1) + f_2(1) \, , \end{align} where for $a>0$ we let \begin{align} f_1(a) &\equiv \int \limits_0^\infty x \operatorname{arccot}(a x) [1-x \operatorname{arccot}(x)] \, \mathrm{d} x \, , \\ f_2(a) &\equiv \int \limits_0^\infty x \operatorname{arccot}(a x) [1-x^2 \operatorname{arccot}^2(x)] \, \mathrm{d} x \, . \\ \end{align}
Computation of $f_1(1)$
Differentiating under the integral sign yields \begin{align} f_1 '(a) &= -\int \limits_0^\infty \frac{x^2}{1+a^2 x^2} [1-x \operatorname{arccot}(x)] \, \mathrm{d} x \\ &= -\frac{1}{a^2}\left[\int \limits_0^\infty [1-x \operatorname{arccot}(x)] \, \mathrm{d} x - \int \limits_0^\infty \frac{1-x \operatorname{arccot}(x)}{1+a^2 x^2} \, \mathrm{d} x \right] \\ &\equiv -\frac{I_1 + g_1 (a)}{a^2} \, . \end{align} We integrate by parts to find \begin{align} I_1 &= \lim_{r \to \infty} \left[r-\frac{r^2}{2} \operatorname{arccot}(r) - \frac{1}{2} \int \limits_0^r \frac{x^2}{1+x^2} \, \mathrm{d} x\right] \\ &= \lim_{r \to \infty} \left[\frac{r-r^2 \operatorname{arccot}(r)}{2}+ \frac{1}{2} \int \limits_0^r \frac{\mathrm{d} x}{1+x^2} \right] \\ &= \frac{\pi}{4} \, . \end{align} Another integration by parts reduces $g_1(a)$ to $$g_1(a) = -\frac{\pi}{2 a} + \frac{1}{2a^2} \int \limits_0^\infty \frac{\ln(1+a^2 x^2)}{1+x^2} \, \mathrm{d} x \, . $$ The remaining integral can be evaluated using differentiation under the integral sign and a partial fraction decomposition. We find $$g_1(a) = -\frac{\pi}{2 a} + \frac{\pi \ln(1+a)}{2a^2} \, . $$ Since $f_1(\infty) = 0$, we can integrate to get \begin{align} f_1(1) &= \int \limits_\infty^1 f_1'(a) \, \mathrm{d} a = \int \limits_1^\infty \left[\frac{\pi}{4a^2} - \frac{\pi}{2 a^3} + \frac{\pi}{2 a^4} \ln(1+a)\right] \, \mathrm{d} a \\ &= \frac{\pi}{3}\ln(2) - \frac{\pi}{12} \, . \end{align}
Computation of $f_2(1)$
In the same manner we can compute \begin{align} f_2 '(a) &= -\int \limits_0^\infty \frac{x^2}{1+a^2 x^2} [1-x^2 \operatorname{arccot}^2(x)] \, \mathrm{d} x \\ &= -\frac{1}{a^2}\left[\int \limits_0^\infty [1-x^2 \operatorname{arccot}^2(x)] \, \mathrm{d} x - \int \limits_0^\infty \frac{1-x^2 \operatorname{arccot}^2(x)}{1+a^2 x^2} \, \mathrm{d} x \right] \\ &\equiv -\frac{I_2 + g_2 (a)}{a^2} \, . \end{align} We now need to integrate by parts twice to evaluate \begin{align} I_2 &= \frac{1}{3} \left[\int \limits_0^\infty \frac{1}{1+x^2} \, \mathrm{d} x + \int \limits_0^\infty \frac{\ln(1+x^2)}{1+x^2} \, \mathrm{d} x + \lim_{r\to\infty} r \{2-r \operatorname{arccot}(r)[1+r\operatorname{arccot}(r)]\}\right] \\ &= \frac{\pi[2 \ln(2)+1]}{6} \, . \end{align} We can write $$ g_2 (a) = - \frac{\pi}{2 a} + \int \limits_0^\infty \frac{x^2 \operatorname{arccot}^2 (x)}{1+a^2 x^2} \, \mathrm{d} x \equiv - \frac{\pi}{2a} + h(a,1) $$ with $$ h(a,b) = \int \limits_0^\infty \frac{x^2 \operatorname{arccot}^2 (b x)}{1+a^2 x^2} \, \mathrm{d} x $$ for $a,b >0$. Differentiation under the integral sign and partial fractions reduce the derivative of $h(a,b)$ with respect to $b$ to known integrals: \begin{align} \partial_b h(a,b) &= - 2 \int \limits_0^\infty \frac{x^3 \operatorname{arccot} (b x)}{(1+a^2 x^2)(1+a^2 x^2)} \, \mathrm{d} x \\ &= \frac{2}{a^2-b^2}\left[\int \limits_0^\infty \frac{x \operatorname{arccot} (b x)}{1+a^2 x^2} \, \mathrm{d} x - \int \limits_0^\infty \frac{x \operatorname{arccot} (b x)}{1+b^2 x^2} \, \mathrm{d} x \right] \\ &= \frac{\pi[\ln(a+b)-\ln(2b)]}{a^2(a^2-b^2)} - \frac{\pi \ln(2)}{a^2 b^2} \, . \end{align} After integrating with respect to $a$ and $b$ and evaluating all elementary integrals we are left with $$f_2(1) = \pi \ln(2) - \frac{\pi}{24} - \pi J \, ,$$ where \begin{align} J &= \frac{1}{2} \int \limits_1^\infty \int \limits_1^\infty \frac{b^{-4} \ln(a) - a^{-4} \ln(b)}{a^2-b^2} \, \mathrm{d} a \, \mathrm{d} b \\ &= \frac{1}{2} \int \limits_0^1 \int \limits_0^1\frac{v^4 \ln(u) - u^4 \ln(v)}{u^2-v^2} \, \mathrm{d} u \, \mathrm{d} v \, . \end{align} This integral can be evaluated by introducing the similar integral $$ K = \frac{1}{2} \int \limits_0^1 \int \limits_0^1\frac{v^4 \ln(v) - u^4 \ln(u)}{u^2-v^2} \, \mathrm{d} u \, \mathrm{d} v \, . $$ We have \begin{align} K &= - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} s} \int \limits_0^1 \int \limits_0^1\frac{u^s - v^s}{u^2-v^2} \, \mathrm{d} u \, \mathrm{d} v ~~ \Bigg \rvert_{s=4} = - \frac{\mathrm{d}}{\mathrm{d} s} \left[\frac{1}{s} \int \limits_0^1 \frac{1-t^s}{1-t^2} \, \mathrm{d} t \right] \\ &= \frac{1}{16} \int \limits_0^1 (1+t^2)\, \mathrm{d} t - \frac{1}{4} \int \limits_0^1 \frac{- \ln(t) t^4}{1-t^2} \, \mathrm{d} t \\ &= \frac{1}{12} - \frac{1}{4} \sum_{k=0}^\infty \frac{1}{(2k+5)^2} = \frac{13}{36} - \frac{\pi^2}{32} \, . \end{align} Since $$ J + K = \frac{1}{2} \int \limits_0^1 \int \limits_0^1 - \ln(u v) (u^2+v^2) \, \mathrm{d} u \, \mathrm{d} v = \frac{4}{9} \, , $$ we conclude that $$ J = J + K - K = \frac{1}{12} + \frac{\pi^2}{32} \, .$$ Therefore we arrive at $$ f_2(1) = \pi \ln(2) - \frac{\pi}{8} - \frac{\pi^3}{32} \, . $$
Final result
We now obtain the value $$ I = f_1(1) + f_2(1) = \frac{4\pi}{3} \ln(2) - \frac{5\pi}{24} - \frac{\pi^3}{32} \, . $$ for the original integral. Note that $I \approx 1.2800035$ is very close but not equal to the rational number $\frac{32}{25} = 1.28$ .