The image shows a double integral problem. The integral is as follows:

$(b) \int_1^{\ln 8} \int_0^{\ln y} e^{x + y} \, dx \, dy$

This is a double integral with respect to $x$ and $y$, where the limits for $x$ range from $0$ to $\ln y$, and the limits for $y$ range from $1$ to $\ln 8$. The integrand is $e^{x + y}$.

Steps to Solve:

1. Integrate with respect to $x$: The first step is to compute the inner integral: $\int_0^{\ln y} e^{x + y} \, dx$ Since $e^{x + y} = e^x \cdot e^y$, we can factor out $e^y$: $e^y \int_0^{\ln y} e^x \, dx$ The integral of $e^x$ with respect to $x$ is simply $e^x$. Evaluating this from $0$ to $\ln y$: $e^y \left[ e^{\ln y} - e^0 \right] = e^y \left( y - 1 \right)$

2. Integrate with respect to $y$: Now we integrate the result with respect to $y$: $\int_1^{\ln 8} e^y (y - 1) \, dy$ This can be broken into two integrals: $\int_1^{\ln 8} y e^y \, dy - \int_1^{\ln 8} e^y \, dy$

   The second term is straightforward: $\int_1^{\ln 8} e^y \, dy = e^y \bigg|_1^{\ln 8} = e^{\ln 8} - e^1 = 8 - e$

   For the first term, we use integration by parts. Let: $u = y, \quad dv = e^y \, dy \quad \Rightarrow \quad du = dy, \quad v = e^y$ Applying integration by parts: $\int y e^y \, dy = y e^y - \int e^y \, dy = y e^y - e^y$ Evaluating this from $1$ to $\ln 8$: $\left( y e^y - e^y \right) \bigg|_1^{\ln 8} = \left( \ln 8 \cdot 8 - 8 \right) - \left( 1 \cdot e - e \right) = 8 \ln 8 - 8 - e + e = 8 \ln 8 - 8$

3. Final Answer: Putting it all together: $(8 \ln 8 - 8) - (8 - e) = 8 \ln 8 - 16 + e$

Thus, the solution to the integral is:

$8 \ln 8 - 16 + e$

Would you like more details or have any questions?

Here are 5 related questions to consider:

1. How does integration by parts work and when should it be applied?
2. Why can $e^{x+y}$ be separated as $e^x \cdot e^y$ in the integrand?
3. What are some strategies to approach complex limits in double integrals?
4. How would you approach this integral if the limits were different, say, from 0 to $y$ instead of $\ln y$?
5. What happens if you reverse the order of integration for this double integral?

Tip: When dealing with exponential integrals, factoring and separating terms can simplify the calculation significantly.