\[ A = \int_{1/2}^{1+\sqrt{5}} \left(1 + 3x - 2x^2 - \frac{1}{x}\right) \, dx \]
\[ A = \left[ x + \frac{3x^2}{2} - \frac{2x^3}{3} - \ln x \right]_{1/2}^{1+\sqrt{5}} \]
\[ A = \left( \left(1 + \sqrt{5}\right) + \frac{3\left(1+\sqrt{5}\right)^2}{2} - \frac{2\left(1+\sqrt{5}\right)^3}{3} - \ln\left(1+\sqrt{5}\right) \right) - \left( \frac{1}{2} + \frac{3\left(1/2\right)^2}{2} - \frac{2\left(1/2\right)^3}{3} - \ln\left(\frac{1}{2}\right) \right) \]
Simplify step-by-step:
\[ A = (1 + \sqrt{5}) + \frac{3}{2}(1+\sqrt{5})^2 - \frac{2}{3}(1+\sqrt{5})^3 - \ln(1+\sqrt{5}) - \left( \frac{1}{2} + \frac{3}{8} - \frac{1}{12} + \ln 2 \right) \]
\[ A = \frac{1}{2} + \sqrt{5} + \frac{3}{2}(1 + 2\sqrt{5} + 5) - \frac{2}{3}(1 + 3\sqrt{5} + 3(5) + 5\sqrt{5}) - \ln(1+\sqrt{5}) - \frac{1}{2} - \frac{3}{8} + \frac{1}{12} - \ln 2 \]
Combine and simplify further:
\[ A = \sqrt{5}\left(1 + \frac{3}{2} - 2\right) + \frac{15}{8} - \frac{4}{3} + \frac{1}{12} - \ln(1+\sqrt{5}) \]
\[ A = \frac{14\sqrt{5}}{24} + \frac{15}{24} - \ln(1+\sqrt{5}) \]
Final answer: \[ A = \frac{14\sqrt{5}}{24} + \frac{15}{24} - \ln(1+\sqrt{5}) \]
The portion of the line \( 4x + 5y = 20 \) in the first quadrant is trisected by the lines \( L_1 \) and \( L_2 \) passing through the origin. The tangent of an angle between the lines \( L_1 \) and \( L_2 \) is: