(a) \( \int_{1}^{3} \int_{1}^{5} \frac{\ln y}{x y} d y \)

To approach the double integral \( \int_{1}^{3} \int_{1}^{5} \frac{\ln y}{x y} d y \), we start by noticing that the integrand can be separated: \[ \frac{\ln y}{x y} = \frac{1}{x} \cdot \frac{\ln y}{y} \] Thus, we can rewrite the double integral as: \[ \int_{1}^{3} \left( \frac{1}{x} \int_{1}^{5} \frac{\ln y}{y} d y \right) dx \] Now, we first need to evaluate the inner integral \( \int_{1}^{5} \frac{\ln y}{y} d y \). This integral can be computed using the substitution \( u = \ln y \), which gives \( du = \frac{1}{y} dy \). When \( y = 1 \), \( u = 0 \), and when \( y = 5 \), \( u = \ln 5 \). Thus, we have: \[ \int_{1}^{5} \frac{\ln y}{y} d y = \int_{0}^{\ln 5} u \, du = \left[ \frac{u^2}{2} \right]_{0}^{\ln 5} = \frac{(\ln 5)^2}{2} \] Putting this back into our expression, we have: \[ \int_{1}^{3} \left( \frac{1}{x} \cdot \frac{(\ln 5)^2}{2} \right) dx = \frac{(\ln 5)^2}{2} \int_{1}^{3} \frac{1}{x} dx \] The integral \( \int_{1}^{3} \frac{1}{x} dx \) evaluates to \( \ln 3 \). Therefore, the entire integral becomes: \[ \frac{(\ln 5)^2}{2} \cdot \ln 3 \] So, the final answer is: \[ \int_{1}^{3} \int_{1}^{5} \frac{\ln y}{x y} d y = \frac{(\ln 5)^2}{2} \ln 3 \]