The function f(x) = cot-1 x + x increases in the interval

The correct answer is (-∞, ∞).

To find the interval in which the given function increases, we need to analyze its first derivative. Let the function be:
$$f\left(x\right)={cot}^{-1}x+x$$

Recall the rule for finding the interval of increase: a function $f\left(x\right)$ is increasing on an interval if its derivative satisfies $f^{\prime }\left(x\right)\ge 0$ for all values of $x$ in that interval (with equality holding at only isolated points).

Let us differentiate $f\left(x\right)$ with respect to $x$:
$$f^{\prime }\left(x\right)=\frac{d}{dx}\left({cot}^{-1}x+x\right)$$

Using standard differentiation formulas, we know that:
$$\frac{d}{dx}\left({cot}^{-1}x\right)=-\frac{1}{1+x^2}$$
and
$$\frac{d}{dx}\left(x\right)=1$$

Substituting these derivatives back, we get:
$$f^{\prime }\left(x\right)=-\frac{1}{1+x^2}+1$$

To simplify this expression, find a common denominator:
$$f^{\prime }\left(x\right)=\frac{-1+\left(1+x^2\right)}{1+x^2}$$
$$f^{\prime }\left(x\right)=\frac{x^2}{1+x^2}$$

Now, let us analyze the sign of $f^{\prime }\left(x\right)$ for all real numbers $x$:
1. For any real number $x$, the numerator $x^2\ge 0$.
2. The denominator $1+x^2\ge 1>0$ is strictly positive for all real values of $x$.

Consequently, the quotient satisfies:
$$f^{\prime }\left(x\right)=\frac{x^2}{1+x^2}\ge 0$$
for all $x\in (-\infty ,\infty )$.

Since the derivative is strictly positive everywhere except at the isolated point $x=0$ (where it is equal to zero), the function $f\left(x\right)$ is strictly increasing on the entire set of real numbers.

Therefore, the function increases in the interval (-∞, ∞).