1. **Problem:** Calculate the integral $$\int \frac{x^4 - 3}{x^2} \, dx$$. 2. **Formula and rules:** Recall that dividing powers of $x$ means subtracting exponents: $$\frac{x^a}{x^b} = x^{a-b}$$. 3. **Simplify the integrand:** $$\frac{x^4 - 3}{x^2} = \frac{x^4}{x^2} - \frac{3}{x^2} = x^{4-2} - 3x^{-2} = x^2 - 3x^{-2}$$ 4. **Rewrite the integral:** $$\int (x^2 - 3x^{-2}) \, dx$$ 5. **Integrate term-by-term:** - For $x^2$, use $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$ when $n \neq -1$. - For $x^{-2}$, similarly. 6. **Calculate each integral:** $$\int x^2 \, dx = \frac{x^{3}}{3}$$ $$\int -3x^{-2} \, dx = -3 \int x^{-2} \, dx = -3 \left( \frac{x^{-1}}{-1} \right) = 3x^{-1}$$ 7. **Combine results:** $$\int \frac{x^4 - 3}{x^2} \, dx = \frac{x^3}{3} + 3x^{-1} + C = \frac{x^3}{3} + \frac{3}{x} + C$$ **Final answer:** $$\boxed{\frac{x^3}{3} + \frac{3}{x} + C}$$