1. **State the problem:** We need to solve the integral $$\int \frac{x^3 + 1}{x^5} \, dx$$. 2. **Rewrite the integrand:** Split the fraction into separate terms: $$\int \frac{x^3}{x^5} + \frac{1}{x^5} \, dx = \int x^{3-5} + x^{-5} \, dx = \int x^{-2} + x^{-5} \, dx$$. 3. **Recall the integration rule:** For any power $n \neq -1$, $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$. 4. **Integrate each term:** - For $x^{-2}$: $$\int x^{-2} \, dx = \frac{x^{-2+1}}{-2+1} = \frac{x^{-1}}{-1} = -x^{-1}$$. - For $x^{-5}$: $$\int x^{-5} \, dx = \frac{x^{-5+1}}{-5+1} = \frac{x^{-4}}{-4} = -\frac{1}{4} x^{-4}$$. 5. **Combine the results:** $$\int \frac{x^3 + 1}{x^5} \, dx = -x^{-1} - \frac{1}{4} x^{-4} + C = -\frac{1}{x} - \frac{1}{4x^4} + C$$. **Final answer:** $$\boxed{-\frac{1}{x} - \frac{1}{4x^4} + C}$$