1. **State the problem:** We need to find the area of the second trapezium under the curve $y=\frac{4}{\ln(x)}$ between two vertical lines where the heights of the trapezium are approximately $2.996$ and $2.621$, and the horizontal distance (base width) between these lines is $\frac{4}{5}$. 2. **Recall the trapezium area formula:** The area $A$ of a trapezium with parallel sides (heights) $h_1$ and $h_2$ and base width $b$ is given by $$A = \frac{1}{2} (h_1 + h_2) \times b$$ 3. **Substitute the given values:** $$h_1 = 2.996, \quad h_2 = 2.621, \quad b = \frac{4}{5} = 0.8$$ 4. **Calculate the sum of the heights:** $$h_1 + h_2 = 2.996 + 2.621 = 5.617$$ 5. **Calculate the area:** $$A = \frac{1}{2} \times 5.617 \times 0.8 = 0.5 \times 5.617 \times 0.8$$ $$A = 2.8085$$ 6. **Round the answer to 3 decimal places:** $$A \approx 2.809$$ **Final answer:** The area of the second trapezium is approximately **2.809**.