If \( \int_{0}^{k} \frac{x}{x^{2}+4} d x=\frac{1}{2} \ln 4 \), where \( k>0 \), then \( k= \)

To solve the integral \( \int_{0}^{k} \frac{x}{x^{2}+4} \, dx = \frac{1}{2} \ln 4 \), we will first need to evaluate the integral on the left-hand side. 1. **Finding the antiderivative**: We can use substitution to evaluate the integral. Let \( u = x^2 + 4 \), then \( du = 2x \, dx \) or \( \frac{1}{2} du = x \, dx \). The limits of integration change as follows: - When \( x = 0 \), \( u = 0^2 + 4 = 4 \). - When \( x = k \), \( u = k^2 + 4 \). Now rewriting the integral in terms of \( u \): \[ \int_{0}^{k} \frac{x}{x^2 + 4} \, dx = \int_{4}^{k^2 + 4} \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} \int_{4}^{k^2 + 4} \frac{1}{u} \, du \] The integral of \( \frac{1}{u} \) is \( \ln|u| \), leading to: \[ \frac{1}{2} \left[ \ln|u| \right]_{4}^{k^2 + 4} = \frac{1}{2} \left( \ln(k^2 + 4) - \ln(4) \right) = \frac{1}{2} \ln\left( \frac{k^2 + 4}{4} \right) \] So we have: \[ \int_{0}^{k} \frac{x}{x^2 + 4} \, dx = \frac{1}{2} \ln\left( \frac{k^2 + 4}{4} \right) \] 2. **Setting up the equation**: We equate this to the given value: \[ \frac{1}{2} \ln\left( \frac{k^2 + 4}{4} \right) = \frac{1}{2} \ln 4 \] Since we have the same factor of \( \frac{1}{2} \), we can multiply both sides by 2: \[ \ln\left( \frac{k^2 + 4}{4} \right) = \ln 4 \] 3. **Exponentiating both sides**: By exponentiating, we eliminate the logarithm: \[ \frac{k^2 + 4}{4} = 4 \] Multiplying through by 4 gives: \[ k^2 + 4 = 16 \] Thus, \[ k^2 = 16 - 4 = 12 \] And taking the square root (noting that \( k > 0 \)) gives: \[ k = \sqrt{12} = 2\sqrt{3} \] Thus, the final answer is: \[ \boxed{2\sqrt{3}} \]