TITLE: Integral $\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$
QUESTION [53 upvotes]: Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$$

REPLY [2 votes]: I just want to give two alternate forms of
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}dx=\pi\ln\frac34-\Im\operatorname{Li}_2\frac{3i}4.$$
First by using $(12)$ from here we get
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}dx=\pi\ln\frac{3}{2}-6\,G+4\,\Im\operatorname{Li}_2\frac{i}{2}+2\,\Im\operatorname{Li}_2\frac{i}{3},$$
where $G$ is the Catalan's constant.
The second one comes from the relationship between inverse tangent integral and Clausen's integral.
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}dx=\pi\ln\left(\frac{3}{4}\right)+\arctan\left(\frac{3}{4}\right)\ln\left(\frac{4}{3}\right)-\frac{1}{2}\Im\left(\operatorname{Li}_2\left(\alpha^2\right)+\operatorname{Li}_2\left(-\frac{1}{\alpha^2}\right)\right),$$
where $\alpha=\frac{1+2i}{2+i}=\frac{4}{5}+\frac{3}{5}i$. Note that $\arctan\frac{3}{4}=\arcsin\frac{3}{5}$.