1. **State the problem:** Calculate the definite integral $$\int_{-1}^5 (1 + 3x) \, dx$$. 2. **Recall the formula:** The integral of a sum is the sum of the integrals, and the integral of $ax^n$ is $\frac{a}{n+1}x^{n+1}$. 3. **Set up the integral:** $$\int_{-1}^5 (1 + 3x) \, dx = \int_{-1}^5 1 \, dx + \int_{-1}^5 3x \, dx$$ 4. **Integrate each term:** $$\int 1 \, dx = x$$ $$\int 3x \, dx = \frac{3}{2}x^2$$ 5. **Evaluate the definite integrals:** $$\left[x\right]_{-1}^5 + \left[\frac{3}{2}x^2\right]_{-1}^5 = (5 - (-1)) + \left(\frac{3}{2} \times 5^2 - \frac{3}{2} \times (-1)^2\right)$$ 6. **Calculate values:** $$5 - (-1) = 6$$ $$\frac{3}{2} \times 25 = 37.5$$ $$\frac{3}{2} \times 1 = 1.5$$ 7. **Combine results:** $$6 + (37.5 - 1.5) = 6 + 36 = 42$$ **Final answer:** $$\int_{-1}^5 (1 + 3x) \, dx = 42$$