If $y=2x+{\cot }^{-1}x+\log \left(\sqrt{1+x^2}-x\right)$ then $y$

$y^{\text{'}}\left(x\right)=2-\frac{1}{1+x^2}+\frac{1}{\sqrt{1+x^2}-x}\frac{d}{dx}\left(\sqrt{1+x^2}-x\right)$

$=2-\frac{1}{1+x^2}+\frac{1}{\sqrt{1+x^2}-x}\times \left(\frac{x}{\sqrt{1+x^2}}-1\right)$

$=2-\frac{1}{1+x^2}-\frac{1}{\sqrt{1+x^2}}\ge 0$

since $1/\left(1+x^2\right)$ and $1/\sqrt{1+x^2}$ are less than or equal to 1 for all $x$. So $f\left(x\right)$ increases on $(-\infty ,\infty )$.