1. **Problem:** Evaluate the definite integral \( \int_1^2 \left( \frac{2}{x^3} + 3x \right) dx \). 2. **Formula and rules:** The definite integral of a sum is the sum of the integrals. Use power rule for integration: \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \) for \( n \neq -1 \). 3. **Step-by-step solution:** \[ \int_1^2 \left( \frac{2}{x^3} + 3x \right) dx = \int_1^2 2x^{-3} dx + \int_1^2 3x dx \] Integrate each term: \[ \int 2x^{-3} dx = 2 \int x^{-3} dx = 2 \cdot \frac{x^{-2}}{-2} = -x^{-2} = -\frac{1}{x^2} \] \[ \int 3x dx = 3 \cdot \frac{x^2}{2} = \frac{3x^2}{2} \] Evaluate from 1 to 2: \[ \left[-\frac{1}{x^2} + \frac{3x^2}{2} \right]_1^2 = \left(-\frac{1}{2^2} + \frac{3 \cdot 2^2}{2} \right) - \left(-\frac{1}{1^2} + \frac{3 \cdot 1^2}{2} \right) \] Calculate values: \[ = \left(-\frac{1}{4} + \frac{3 \cdot 4}{2} \right) - \left(-1 + \frac{3}{2} \right) = \left(-\frac{1}{4} + 6 \right) - \left(-1 + 1.5 \right) \] \[ = \left( \frac{23}{4} \right) - \left( \frac{1}{2} \right) = \frac{23}{4} - \frac{2}{4} = \frac{21}{4} = 5.25 \] **Final answer:** \( \int_1^2 \left( \frac{2}{x^3} + 3x \right) dx = \frac{21}{4} \).