$f(x)=2 x-\tan ^{-1} x-\log (x+\sqrt{x^2+1}) \text { is monotonically increasing, when }$

The function is given by $f(x) = 2x - \tan^{-1} x - \log(x + \sqrt{x^2 + 1})$.

To determine the interval where $f(x)$ is monotonically increasing, we find its derivative $f'(x)$ with respect to $x$.

$f'(x) = \dfrac{d}{dx}(2x) - \dfrac{d}{dx}(\tan^{-1} x) - \dfrac{d}{dx}(\log(x + \sqrt{x^2 + 1}))$.

The derivative of $\tan^{-1} x$ is $\dfrac{1}{1 + x^2}$.

The derivative of $\log(x + \sqrt{x^2 + 1})$ is $\dfrac{1}{x + \sqrt{x^2 + 1}} \times (1 + \dfrac{2x}{2\sqrt{x^2 + 1}}) = \dfrac{1}{x + \sqrt{x^2 + 1}} \times (\dfrac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}}) = \dfrac{1}{\sqrt{x^2 + 1}}$.

Thus, $f'(x) = 2 - \dfrac{1}{1 + x^2} - \dfrac{1}{\sqrt{x^2 + 1}}$.

Let $t = \dfrac{1}{\sqrt{x^2 + 1}}$. Since $x^2 \ge 0$, we have $x^2 + 1 \ge 1$, so $0 < t \le 1$.

Then $\dfrac{1}{1 + x^2} = t^2$.

$f'(x) = 2 - t^2 - t = -(t^2 + t - 2) = -(t + 2)(t - 1) = (1 - t)(t + 2)$.

Since $0 < t \le 1$, $1 - t \ge 0$ and $t + 2 > 0$.

Therefore, $f'(x) \ge 0$ for all $x \in R$. Specifically, $f'(x) = 0$ only at $t = 1$, which corresponds to $x = 0$.

Since $f'(x) > 0$ for $x \neq 0$ and $f(x)$ is continuous, $f(x)$ is monotonically increasing for all $x \in R$.

Answer: $x \in R$