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?

Consider the following integral:
$$I(a)=\int_0^\infty \frac{\operatorname{arccot}(a\sqrt{1+x^2})}{\sqrt{1+x^2}}dx\Rightarrow I'(a)=-\int_0^\infty \frac{1}{1+a^2+a^2x^2}dx$$
$$=-\frac{1}{a\sqrt{1+a^2}}\arctan\left(\frac{ax}{\sqrt{1+a^2}}\right)\bigg|_0^\infty=-\frac{\pi}{2}\frac{1}{a\sqrt{1+a^2}} $$
Since $I(\infty)=0$ and we are looking for $I(1)$ we have:
$$I(1)=-(I(\infty)-I(1))=\frac{\pi}{2}\int_1^\infty \frac{1}{a\sqrt{1+a^2}}da\overset{a=\frac{1}{x}}=\frac{\pi}{2}\int_0^1 \frac{1}{\sqrt{1+x^2}}dx$$
$$\Rightarrow \boxed{\int_0^\infty \frac{\operatorname{arccot}(\sqrt{1+x^2})}{\sqrt{1+x^2}}dx=\frac{\pi}{2}\ln(1+\sqrt 2)}$$
