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?

Let 
$$I=\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}}dx =\int_{0}^{\infty} \frac{\tan^{-1}(1/\sqrt{(1+x^2})}{\sqrt{1+x^2}} dx$$
Let us use the integral representation of $$\frac{\tan^{-1}(1/\sqrt{1+x^2})}{\sqrt{1+x^2}}=\int_{0}^{1} \frac{dt}{1+x^2+t^2}$$
$$I=\int_{0}^{\infty} \int_{0}^{1} \frac {dt dx}{1+x^2+t^2}=\int_{0}^{1}\frac{dt}{\sqrt{1+t^2}} \tan^{-1} \left.\frac{x}{\sqrt{1+t^2}}\right|_{0}^{\infty} =\frac{\pi}{2} \ln (1+\sqrt{2}).$$
