1. **State the problem:** We need to find the value of $k > 0$ such that $$\int_0^k \frac{x}{x^2 + 4} \, dx = \frac{1}{2} \ln 4.$$\n\n2. **Recall the formula and substitution:** The integral involves a rational function where the numerator is the derivative of the denominator's inner function. We use substitution: let $$u = x^2 + 4,$$ then $$du = 2x \, dx \implies x \, dx = \frac{du}{2}.$$\n\n3. **Rewrite the integral:**\n$$\int_0^k \frac{x}{x^2 + 4} \, dx = \int_{u=4}^{u=k^2 + 4} \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \int_4^{k^2 + 4} \frac{1}{u} \, du.$$\n\n4. **Integrate:**\n$$\frac{1}{2} \int_4^{k^2 + 4} \frac{1}{u} \, du = \frac{1}{2} \left[ \ln u \right]_4^{k^2 + 4} = \frac{1}{2} \left( \ln(k^2 + 4) - \ln 4 \right) = \frac{1}{2} \ln \frac{k^2 + 4}{4}.$$\n\n5. **Set equal to given value and solve for $k$:**\n$$\frac{1}{2} \ln \frac{k^2 + 4}{4} = \frac{1}{2} \ln 4.$$\nMultiply both sides by 2:\n$$\ln \frac{k^2 + 4}{4} = \ln 4.$$\nExponentiate both sides to remove the logarithm:\n$$\frac{k^2 + 4}{4} = 4.$$\nMultiply both sides by 4:\n$$k^2 + 4 = 16.$$\nSubtract 4 from both sides:\n$$k^2 = 12.$$\nTake the positive square root (since $k > 0$):\n$$k = \sqrt{12} = 2\sqrt{3}.$$\n\n6. **Final answer:**\n$$k = \sqrt{12}.$$\n\n**Therefore, the correct choice is D: $\sqrt{12}$.