1. **State the problem:** We want to solve the integral $$\int (4x^2 - 12x + 9)^{\frac{2}{3}} \, dx$$ using substitution. 2. **Identify substitution:** Notice the expression inside the parentheses is a quadratic. Let's try to simplify it by completing the square: $$4x^2 - 12x + 9 = (2x)^2 - 2 \cdot 2x \cdot 3 + 3^2 = (2x - 3)^2$$ 3. **Rewrite the integral:** $$\int (4x^2 - 12x + 9)^{\frac{2}{3}} \, dx = \int ((2x - 3)^2)^{\frac{2}{3}} \, dx = \int (2x - 3)^{\frac{4}{3}} \, dx$$ 4. **Use substitution:** Let $$u = 2x - 3$$ Then, $$\frac{du}{dx} = 2 \implies dx = \frac{du}{2}$$ 5. **Rewrite integral in terms of $u$:** $$\int (2x - 3)^{\frac{4}{3}} \, dx = \int u^{\frac{4}{3}} \cdot \frac{du}{2} = \frac{1}{2} \int u^{\frac{4}{3}} \, du$$ 6. **Integrate:** Use the power rule for integrals: $$\int u^n \, du = \frac{u^{n+1}}{n+1} + C$$ Here, $n = \frac{4}{3}$, so $$\int u^{\frac{4}{3}} \, du = \frac{u^{\frac{4}{3} + 1}}{\frac{4}{3} + 1} + C = \frac{u^{\frac{7}{3}}}{\frac{7}{3}} + C = \frac{3}{7} u^{\frac{7}{3}} + C$$ 7. **Substitute back:** $$\frac{1}{2} \int u^{\frac{4}{3}} \, du = \frac{1}{2} \cdot \frac{3}{7} u^{\frac{7}{3}} + C = \frac{3}{14} (2x - 3)^{\frac{7}{3}} + C$$ **Final answer:** $$\int (4x^2 - 12x + 9)^{\frac{2}{3}} \, dx = \frac{3}{14} (2x - 3)^{\frac{7}{3}} + C$$