We have
$$
\begin{aligned}
&y=2 x+\cot ^{-1} x+\log \left[\sqrt{1+x^2}-x\right] \\
&\Rightarrow \frac{d y}{d x}=2-\frac{1}{1+x^2}+\frac{1}{\sqrt{1+x^2}-x} \times\left[\frac{x}{\sqrt{1+x^2}}-1\right]
\end{aligned}
$$
$$
=\frac{2 x^2+1}{1+x^2}-\frac{1}{\sqrt{1+x^2}}=\frac{\left(2 x^2+1\right)-\sqrt{1+x^2}}{1+x^2}
$$
Now, $\frac{\mathrm{dy}}{\mathrm{dx}} \geq 0$
$$
\begin{aligned}
&\Rightarrow\left(2 x^2+1\right)-\sqrt{1+x^2} \geq 0 \\
&\Rightarrow\left(2 x^2+1\right)^2 \geq 1+x^2 \Rightarrow 4 x^4+3 x^2 \geq 0
\end{aligned}
$$
Which is true for all real values of $x$.
$\therefore y$ increases for all real values of $x$.