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

$f(x)=x+\cot ^{-1} x$

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 ∈ (-∞, ∞)