Concept: A function is increasing if its derivative is positive. Compute \(f'(x)\), simplify using hyperbolic-like substitutions, and show it's always non-negative.
Step 1: \(f(x) = 2x + \cot^{-1}x + \log(\sqrt{1+x^2}-x)\). Note \(\sqrt{1+x^2}-x = (\sqrt{1+x^2}+x)^{-1}\), so \(\log(\sqrt{1+x^2}-x) = -\log(\sqrt{1+x^2}+x)\).
Step 2: Derivative: \(f'(x) = 2 - \frac{1}{1+x^2} - \frac{1}{\sqrt{1+x^2}}\). Let \(t=\sqrt{1+x^2} \ge 1\). Then \(f'(x) = 2 - 1/t - 1/t^2 = \frac{2t^2 - t - 1}{t^2} = \frac{(2t+1)(t-1)}{t^2}\).
Step 3: For all \(x\), \(t\ge 1 \Rightarrow t-1 \ge 0, 2t+1 > 0, t^2 > 0\). So \(f'(x) \ge 0\), and \(>0\) for \(x \neq 0\). Thus f increases on \((-\infty,\infty)\).
Step 4: Write the final answer. \(\boxed{\text{The function increases on }(-\infty,\infty).}\)