Find \( \int \frac{\sqrt{2+\ln z}}{z} d z \) Answer: \( \square \)

To solve the integral \( \int \frac{\sqrt{2+\ln z}}{z} d z \), we can use the substitution \( u = \ln z \). This means that \( du = \frac{1}{z} dz \). Consequently, the variable \( z \) becomes \( e^u \) and the bounds change accordingly with the substitution. After the substitution, the integral transforms to: \[ \int \sqrt{2 + u} \, du \] Next, we proceed to find the antiderivative. The integral \( \int \sqrt{2 + u} \, du \) can be solved using a standard power rule integral. Letting \( v = 2 + u \), we have \( dv = du \). Now our integral is: \[ \int \sqrt{v} \, dv = \frac{2}{3} v^{3/2} + C = \frac{2}{3} (2 + u)^{3/2} + C \] Substituting back for \( u \) yields: \[ \frac{2}{3} (2 + \ln z)^{3/2} + C \] So, the final answer is \[ \int \frac{\sqrt{2+\ln z}}{z} \, dz = \frac{2}{3} (2 + \ln z)^{3/2} + C \]