1. **State the problem:** We need to evaluate the integral $$\int \frac{\sqrt{x} - x + 5}{x^4} \, dx$$. 2. **Rewrite the integrand:** Express all terms with powers of $x$ to simplify integration. Recall that $\sqrt{x} = x^{\frac{1}{2}}$. So, $$\frac{\sqrt{x} - x + 5}{x^4} = \frac{x^{\frac{1}{2}}}{x^4} - \frac{x}{x^4} + \frac{5}{x^4} = x^{\frac{1}{2} - 4} - x^{1 - 4} + 5x^{-4} = x^{-\frac{7}{2}} - x^{-3} + 5x^{-4}.$$ 3. **Set up the integral with simplified terms:** $$\int \left(x^{-\frac{7}{2}} - x^{-3} + 5x^{-4}\right) dx = \int x^{-\frac{7}{2}} dx - \int x^{-3} dx + 5 \int x^{-4} dx.$$ 4. **Recall the power rule for integration:** For $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, provided $n \neq -1$. 5. **Integrate each term:** - For $\int x^{-\frac{7}{2}} dx$: $$n = -\frac{7}{2}, \quad n+1 = -\frac{7}{2} + 1 = -\frac{5}{2}.$$ So, $$\int x^{-\frac{7}{2}} dx = \frac{x^{-\frac{5}{2}}}{-\frac{5}{2}} = -\frac{2}{5} x^{-\frac{5}{2}}.$$ - For $\int x^{-3} dx$: $$n = -3, \quad n+1 = -2.$$ So, $$\int x^{-3} dx = \frac{x^{-2}}{-2} = -\frac{1}{2} x^{-2}.$$ - For $5 \int x^{-4} dx$: $$n = -4, \quad n+1 = -3.$$ So, $$5 \int x^{-4} dx = 5 \cdot \frac{x^{-3}}{-3} = -\frac{5}{3} x^{-3}.$$ 6. **Combine all results:** $$\int \frac{\sqrt{x} - x + 5}{x^4} dx = -\frac{2}{5} x^{-\frac{5}{2}} - \frac{1}{2} x^{-2} - \frac{5}{3} x^{-3} + C.$$ 7. **Final answer:** $$\boxed{-\frac{2}{5} x^{-\frac{5}{2}} - \frac{1}{2} x^{-2} - \frac{5}{3} x^{-3} + C}.$$