Integral $\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$ Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$$

By putting $x=\sinh(z)^2$ we have:
$$ I = \int_{0}^{+\infty}2\tanh z\operatorname{arccot}\left(\sinh z-2\cosh z\right)dz; $$
integrating by parts we have:
$$ I = \int_{0}^{+\infty} 2\frac{-1+2\tanh z}{5\cosh z-4\sinh x}\log\left(\cosh z\right)dz;$$
now putting $z=\log t$ we get:
$$ I = 4\int_{1}^{+\infty} \frac{t^2-3}{(t^2+1)(t^2+9)}\log\left(\frac{t+t^{-1}}{2}\right)dt.$$
By putting $y=\frac{2}{t+t^{-1}}$ we can split the integral in two parts, i.e.
$$ I_1 = 24\int_{0}^{1}\frac{\log x}{9+16x^2} dx,$$
$$ I_2 = -4\int_{0}^{1}\frac{(3-8x^2)\log x}{(9+16x^2)\sqrt{1-x^2}}dx.$$
By considering the Taylor series of $\frac{1}{9+16x^2}$ and integrating term-by-term we get $I_1=\Im\operatorname{Li}_2\frac{3i}{4}$.
The really mysterious thing is that $I_2=\pi\log\frac{3}{4}$. Not so mysterious, anyway. We have that:
$$\int_{0}^{1}\frac{\log x}{\sqrt{1-x^2}}=\int_{0}^{\pi/2}\log\cos x\, dx=-\frac{1}{2}\int_{0}^{+\infty}\frac{\log(1+t^2)}{1+t^2}=-\frac{\pi}{2}\log 2,$$
(through $x=\arctan t$) in virtue of the magic formula:
$$\int_{\mathbb{R}}\frac{\log\left(A^2 x^2 + B^2\right)}{x^2+1}dx=2\pi\log(A+B),\tag{1}$$
that can be proved through standard complex-analytic techniques. So we have only to deal with:
$$\int_{0}^{1}\frac{\log x}{(16x^2+9)\sqrt{1-x^2}}dx=\int_{0}^{\pi/2}\frac{\log\cos\theta}{16\cos^2\theta+9}d\theta,$$
or, by putting $\theta=\arctan t$,
$$\int_{0}^{\infty}\frac{\log(t^2+1)}{25+9t^2}dt=\frac{1}{15}\int_{0}^{+\infty}\frac{\log(25t^2+9)-\log(9)}{t^2+1}dt,$$
and the magic formula tell us that the last expression is equal to $\frac{1}{15}\left(\pi\log 8-\pi\log 3\right)$, completing the proof.
