Answer
$[-3,1)\cup(1,\infty)$
Work Step by Step
The value inside the square root sign must be non-negative.
So, the function of $k$ is defined when $x+3\geq 0$ or $x\geq -3$.
The denominator of $k(x)$ is not allowed to be zero.
So, the function of $k$ is also defined when $x\neq 1$.
Therefore, the domain of $k$ is
$D_k=\{x|x\geq -3\vee x\neq 1\}$
