Step 1: Factor the denominator:
Write $x + \sqrt{x} = \sqrt{x}\,(\sqrt{x}+1)$.
Step 2: Use a different substitution:
Let $u = 1+\sqrt{x}$. Then $du = \frac{dx}{2\sqrt{x}}$, so $\frac{dx}{\sqrt{x}} = 2\,du$.
The limits change: $x=0$ gives $u=1$, and $x=4$ gives $u=3$.
Step 3: Integrate:
\[ \int_{0}^{4}\frac{dx}{\sqrt{x}(1+\sqrt{x})} = \int_{1}^{3}\frac{2\,du}{u} = 2\ln u\,\Big|_{1}^{3} = 2\ln 3 \]
Step 4: Read off k:
Since the integral equals $k\log 3$, we get $k=2$.
A quick numeric check: the integral is $2\ln 3 \approx 2.197$, and $4\ln 3$ would be too large.
Final Answer:
Option (D) is correct.
\[ \boxed{2} \]