The function $f(x)=x+\cot ^{-1} x$ is increasing in the interval :

$(-\infty, \infty)$ $(-1, \infty)$ $(0, \infty)$ $(1, \infty)$

$(-\infty, \infty)$

$f(x)=x+\cot ^{-1} x$ so differentiating f(x) w.r.t x $f'(x) = 1 - \frac{1}{1+x^2} = \frac{1+x^2-1}{1+x^2}$ so  $f'(x) = \frac{x^2}{1+x^2}$ $x^2>0$ so  $\frac{x^2}{1+x^2} > 0$ ⇒ f(x) is always increasing on R so f(x) is increasing for x ∈ (-∞, ∞)