Step 1: Recall the test.
A function is increasing wherever its derivative is positive (and is non-decreasing where the derivative is at least zero). So we differentiate $f(x)=\cot^{-1}x+x$.
Step 2: Differentiate each part.
The derivative of $\cot^{-1}x$ is $-\frac{1}{1+x^2}$, and the derivative of $x$ is $1$.
Step 3: Combine.
So $f'(x)=1-\frac{1}{1+x^2}$.
Step 4: Put over a common denominator.
$f'(x)=\frac{(1+x^2)-1}{1+x^2}=\frac{x^2}{1+x^2}$.
Step 5: Check the sign.
The top $x^2$ is never negative and the bottom $1+x^2$ is always positive, so $f'(x)\ge 0$ for every real $x$. It is zero only at $x=0$.
Step 6: State the interval.
Since the derivative stays non-negative everywhere, the function increases on the whole real line $(-\infty,\infty)$, which is option (A).
\[ \boxed{\,(-\infty,\infty)\,} \]