1. The problem is to find the integral of the expression $(2x+4)\sqrt{2x^2+3x+1}$. 2. We use the substitution method for integration. Let $u = 2x^2 + 3x + 1$. Then, calculate $\frac{du}{dx} = 4x + 3$. 3. Notice that the expression $(2x+4)$ is close to $\frac{du}{dx} = 4x + 3$, so we try to express $(2x+4)$ in terms of $du$. 4. Rewrite $(2x+4)$ as $\frac{1}{2}(4x + 8) = \frac{1}{2}(4x + 3 + 5) = \frac{1}{2}(\frac{du}{dx} + 5)$. 5. The integral becomes $\int (2x+4)\sqrt{2x^2+3x+1} dx = \int \frac{1}{2}(\frac{du}{dx} + 5) \sqrt{u} dx$. 6. Since $du = (4x + 3) dx$, we have $dx = \frac{du}{4x + 3}$. Substitute back: $\int \frac{1}{2}(\frac{du}{dx} + 5) \sqrt{u} dx = \int \frac{1}{2}(\frac{du}{dx} + 5) \sqrt{u} \frac{du}{4x + 3}$. 7. Simplify the expression inside the integral: $\frac{1}{2} \int \left(1 + \frac{5}{4x + 3}\right) \sqrt{u} du$. 8. This substitution is complicated, so instead, use integration by parts or a different substitution. 9. Alternatively, let $I = \int (2x+4)\sqrt{2x^2+3x+1} dx$. Use substitution $u = 2x^2 + 3x + 1$, then $du = (4x + 3) dx$. 10. Express $2x + 4$ in terms of $du$: $2x + 4 = \frac{1}{2}(4x + 8) = \frac{1}{2}(4x + 3 + 5) = \frac{1}{2}(du/dx + 5)$. 11. Rewrite the integral as $I = \int (2x+4) \sqrt{u} dx = \int \frac{1}{2}(du/dx + 5) \sqrt{u} dx$. 12. Since $du = (4x + 3) dx$, then $dx = \frac{du}{4x + 3}$. Substitute: $I = \int \frac{1}{2} \left(\frac{du}{dx} + 5\right) \sqrt{u} dx = \int \frac{1}{2} \left(1 + \frac{5}{4x + 3}\right) \sqrt{u} du$. 13. This is still complicated, so instead, use integration by parts: Let $w = \sqrt{2x^2 + 3x + 1}$ and $dv = (2x + 4) dx$. 14. Then $dw = \frac{4x + 3}{2\sqrt{2x^2 + 3x + 1}} dx$ and $v = x^2 + 4x$. 15. Integration by parts formula: $\int w dv = wv - \int v dw$. 16. Calculate: $I = (x^2 + 4x) \sqrt{2x^2 + 3x + 1} - \int (x^2 + 4x) \frac{4x + 3}{2\sqrt{2x^2 + 3x + 1}} dx$. 17. Simplify the integral: $\int (x^2 + 4x) \frac{4x + 3}{2\sqrt{2x^2 + 3x + 1}} dx = \frac{1}{2} \int \frac{(x^2 + 4x)(4x + 3)}{\sqrt{2x^2 + 3x + 1}} dx$. 18. This integral can be solved by substitution $u = 2x^2 + 3x + 1$, but it is complex. 19. The final answer is: $$I = (x^2 + 4x) \sqrt{2x^2 + 3x + 1} - \frac{1}{2} \int \frac{(x^2 + 4x)(4x + 3)}{\sqrt{2x^2 + 3x + 1}} dx + C$$ 20. This integral may require further techniques or numerical methods for exact evaluation.