# Trying to solve $\int_{0}^{\infty}\frac{\operatorname{arccot}\left(\sqrt{1+x}+\sqrt{2+x}\right)}{1+x} dx$

$$I=\large \int_{0}^{\infty}\frac{\operatorname{arccot}\left(\sqrt{1+x}+\sqrt{2+x}\right)}{1+x}\mathrm dx$$

$\large u=\sqrt{1+x}$

$\large \mathrm du=\frac{1}{2\sqrt{1+x}}\mathrm dx$

$$I=2\large \int_{1}^{\infty}\frac{\operatorname{arccot}\left(u+\sqrt{1+u^2}\right)}{u}\mathrm du$$

$$I=\large 2\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}\int_{1}^{\infty}\frac{\mathrm du}{u(u+\sqrt{1+u})^{2k+1}}$$

Applying partial fraction decomposition would be very long.

How would we evaluate $I$?

• The thing inside $\text{arccot}$ should be $u+\sqrt{u^2+1}$. – Szeto Jul 27 '18 at 9:28
• Also, differentiation under the integral sign might be a good option. – Szeto Jul 27 '18 at 9:34

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 .