Step 1: Convert to a product inequality:
$\dfrac{x}{1+x} \ge 0$ is the same as $x(1+x) \ge 0$, provided $x \ne -1$.
Step 2: Solve the quadratic inequality:
$x(x+1) \ge 0$ has roots $0$ and $-1$ and opens upward, so it holds for $x \le -1$ or $x \ge 0$.
Step 3: Remove the bad point:
At $x = -1$ the denominator of the original fraction is zero, so exclude it. The domain is $(-\infty,-1)\cup[0,\infty)$.
This is option (A).
Final Answer:
Option (A).
\[ \boxed{(-\infty,-1)\cup[0,\infty) \text{ (A)}} \]