$$I=\large \int_{0}^{\infty}\frac{\operatorname{arccot}\left(\sqrt{1+x}+\sqrt{2+x}\right)}{1+x}\mathrm dx$$
$\large u=\sqrt{1+x}$
$\large \mathrm du=\frac{1}{2\sqrt{1+x}}\mathrm dx$
$$I=2\large \int_{1}^{\infty}\frac{\operatorname{arccot}\left(u+\sqrt{1+u^2}\right)}{u}\mathrm du$$
$$I=\large 2\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}\int_{1}^{\infty}\frac{\mathrm du}{u(u+\sqrt{1+u})^{2k+1}}$$
Applying partial fraction decomposition would be very long.
How would we evaluate $I$?
Please first do what Szeto has mentioned, it's $u+\sqrt{u^2+1}$ and not $1+u+u^2$.
Then use $\,u:=\cot t\,$ and $\,\displaystyle\cot\frac{t}{2}=\cot t + \sqrt{1+\cot^2 t}\,$ for $\,0<t<\pi\,$.
After some small manipulations (e.g. $\,x:=2t\,$) you will get
$\displaystyle I=\frac{1}{2}\int\limits_0^{\pi/2} \frac{x}{\sin x}dx = G\enspace$ where $\,G\,$ is the Catalan constant .