<s>[INST] The integral $\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}}dx =\frac{\pi}{2}\ln (1+\sqrt{2})$ At Mathematica the numerical value of the integral 
$$\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}} dx$$
equals 1.3844.., which is nothing but $\frac{\pi}{2}\ln (1+\sqrt{2})=z$. Also, one of its transformed forms is evaluated to be  $z$ by Mathematica. The question is: How to do it by hand?
 [/INST] Here we will address your integral:
\begin{equation}
I = \int_0^\infty \frac{\cot^{-1}\left(\sqrt{1 + x^2} \right)}{\sqrt{1 + x^2}}\:dx
\end{equation}
We first let $x = \tan(s)$:
\begin{align}
I &= \int_0^\frac{\pi}{2} \frac{\cot^{-1}\left(\sqrt{1 + \tan^2(s)} \right)}{\sqrt{1 + \tan^2(s)}}\cdot \sec^2(s)\:ds = \int_0^\frac{\pi}{2} \sec(s)\cot^{-1}\left(\sec(s)\right)\:ds \nonumber \\
&= \int_0^\frac{\pi}{2} \frac{\arctan\left(\cos(s)\right)}{\cos(s)}\:ds
\end{align}
Here I will now employ Feynman's Trick and introduce the following function:
\begin{equation}
J(t) = \int_0^\frac{\pi}{2} \frac{\arctan(t\cos(s))}{\cos(s)}\:ds
\end{equation}
Here $I = J(1)$ and $J(0) = 0$. ,By Leibniz's Integral Rule we now differentiate under the curve with respect to $t$:
\begin{align}
J'(t) &= \int_0^\frac{\pi}{2} \frac{1}{t^2\cos^2(s) + 1}\cdot \cos(s) \cdot \frac{1}{\cos(s)}\:ds = \int_0^\frac{\pi}{2}\frac{1}{t^2\cos^2(s) + 1}\:ds \nonumber \\
&= \left[ \frac{1}{\sqrt{t^2 + 1}}\arctan\left(\frac{\tan(s)}{\sqrt{t^2 + 1}}\right) \right]_0^\frac{\pi}{2} = \frac{\pi}{2}\frac{1}{\sqrt{t^2 + 1}}
\end{align}
Thus, 
\begin{equation}
J(t) = \frac{\pi}{2} \int \frac{1}{\sqrt{t^2 + 1}}\:dt = \frac{\pi}{2}\sinh^{-1}(t) + C
\end{equation}
Where $C$ is the constant of integration. To resolve $C$ we use our condition $J(0) = 0$:
\begin{equation}
J(0) = 0 = \frac{\pi}{2}\sinh^{-1}(0) + C \rightarrow C = 0
\end{equation} 
Thus, 
\begin{equation}
 J(t) = \frac{\pi}{2}\sinh^{-1}(t)
\end{equation}
We now may resolve $I$
\begin{equation}
I = J(1) = \frac{\pi}{2}\sinh^{-1}(1)
\end{equation}
Noting that 
\begin{equation}
 \sinh^{-1}(t) = \ln\left|t + \sqrt{t^2 + 1}\right|
\end{equation}
We see that another representation for $I$ is given by:
\begin{equation}
I = \frac{\pi}{2}\ln(1 + \sqrt{2})
\end{equation}
</s>