1. **State the problem:** We need to evaluate the integral $$\int \frac{(x^2 + 2)(3x^2 + 2x)}{\sqrt[3]{x^3 + 2x + 1}} \, dx.$$\n\n2. **Rewrite the integral:** The numerator is the product of two polynomials, and the denominator is the cube root of a cubic polynomial. Let's denote $$u = x^3 + 2x + 1.$$ Then the denominator is $$u^{1/3}.$$\n\n3. **Find the derivative of $u$:**\n$$\frac{du}{dx} = 3x^2 + 2.$$\n\n4. **Check if numerator relates to $u'$:** The numerator is $$(x^2 + 2)(3x^2 + 2x).$$ Let's expand it:\n$$ (x^2 + 2)(3x^2 + 2x) = 3x^4 + 2x^3 + 6x^2 + 4x.$$\n\n5. **Express numerator in terms of $x$ and $u'$:** We have $$u' = 3x^2 + 2,$$ but the numerator is degree 4, so let's try to write numerator as $$(3x^2 + 2)(x^2 + 2x)$$ or similar. Actually, numerator is $$(x^2 + 2)(3x^2 + 2x)$$ as given.\n\n6. **Try substitution:** Let $$u = x^3 + 2x + 1,$$ then $$du = (3x^2 + 2) dx.$$\n\n7. **Rewrite numerator:** Notice numerator is $$(x^2 + 2)(3x^2 + 2x) = (3x^2 + 2)(x^2 + 2x) - 4x.$$ This is not straightforward, so try to express numerator as $$f(x) du/dx$$ for some $f(x)$.\n\n8. **Try substitution $t = u^{2/3}$:** Then $$dt = \frac{2}{3} u^{-1/3} du = \frac{2}{3} u^{-1/3} (3x^2 + 2) dx.$$\n\n9. **Rewrite integral:**\n$$\int \frac{(x^2 + 2)(3x^2 + 2x)}{u^{1/3}} dx = \int (x^2 + 2)(3x^2 + 2x) u^{-1/3} dx.$$\n\n10. **Try to express $(x^2 + 2)(3x^2 + 2x) dx$ in terms of $du$ and $u$:** Since $$du = (3x^2 + 2) dx,$$ but numerator has $3x^2 + 2x$, not $3x^2 + 2$.\n\n11. **Try to factor numerator differently:**\n$$ (x^2 + 2)(3x^2 + 2x) = 3x^4 + 2x^3 + 6x^2 + 4x.$$\n\n12. **Try substitution $v = x^3 + 2x + 1$ and express numerator in terms of $v$ and $dv/dx$:**\n$$dv/dx = 3x^2 + 2,$$ but numerator has $3x^2 + 2x$, so no direct match.\n\n13. **Try polynomial division or rewrite numerator:**\nRewrite numerator as $$3x^4 + 2x^3 + 6x^2 + 4x = x(3x^3 + 2x^2 + 6x + 4).$$\n\n14. **Try substitution $w = x^3 + 2x + 1$ and express numerator as $x w'$:**\n$$w' = 3x^2 + 2,$$ so $$x w' = x(3x^2 + 2) = 3x^3 + 2x,$$ which is part of numerator but not all.\n\n15. **Try to split numerator:**\n$$3x^4 + 2x^3 + 6x^2 + 4x = x(3x^3 + 2x^2 + 6x + 4) = x(3x^3 + 2x^2) + x(6x + 4).$$\n\n16. **Try substitution $z = (x^3 + 2x + 1)^{2/3}$:**\nThen $$dz = \frac{2}{3} (x^3 + 2x + 1)^{-1/3} (3x^2 + 2) dx.$$\n\n17. **Rewrite integral in terms of $z$:**\nWe want to express integral as $$\int (x^2 + 2)(3x^2 + 2x) u^{-1/3} dx = \int f(x) dz.$$\n\n18. **Try to express $f(x) dx$ in terms of $dz$:**\nFrom step 16, $$dz = \frac{2}{3} u^{-1/3} (3x^2 + 2) dx,$$ so\n$$u^{-1/3} dx = \frac{3}{2} \frac{dz}{3x^2 + 2}.$$\n\n19. **Rewrite integral:**\n$$\int (x^2 + 2)(3x^2 + 2x) u^{-1/3} dx = \int (x^2 + 2)(3x^2 + 2x) \cdot u^{-1/3} dx = \int (x^2 + 2)(3x^2 + 2x) \cdot \frac{3}{2} \frac{dz}{3x^2 + 2} = \frac{3}{2} \int (x^2 + 2) \frac{3x^2 + 2x}{3x^2 + 2} dz.$$\n\n20. **Simplify the fraction inside integral:**\n$$\frac{3x^2 + 2x}{3x^2 + 2} = 1 - \frac{2 - 2x}{3x^2 + 2}$$ but this is complicated. Instead, try to approximate or check if numerator equals denominator times something.\n\n21. **Try polynomial division:**\nDivide numerator $3x^2 + 2x$ by denominator $3x^2 + 2$:\n$$\frac{3x^2 + 2x}{3x^2 + 2} = 1 + \frac{2x - 2}{3x^2 + 2}.$$\n\n22. **Rewrite integral:**\n$$\frac{3}{2} \int (x^2 + 2) \left(1 + \frac{2x - 2}{3x^2 + 2}\right) dz = \frac{3}{2} \int (x^2 + 2) dz + \frac{3}{2} \int (x^2 + 2) \frac{2x - 2}{3x^2 + 2} dz.$$\n\n23. **Since $z$ is function of $x$, this is complicated, so try direct substitution:**\nTry substitution $$t = (x^3 + 2x + 1)^{1/3} = u^{1/3}.$$ Then $$t^3 = u,$$ and $$dt = \frac{1}{3} u^{-2/3} du = \frac{1}{3} u^{-2/3} (3x^2 + 2) dx.$$\n\n24. **Rewrite integral in terms of $t$:**\n$$\int \frac{(x^2 + 2)(3x^2 + 2x)}{u^{1/3}} dx = \int (x^2 + 2)(3x^2 + 2x) t^{-1} dx.$$\n\n25. **Express $dx$ in terms of $dt$:**\nFrom step 23, $$du = (3x^2 + 2) dx,$$ so $$dx = \frac{du}{3x^2 + 2}.$$\n\n26. **Rewrite integral:**\n$$\int (x^2 + 2)(3x^2 + 2x) t^{-1} dx = \int (x^2 + 2)(3x^2 + 2x) t^{-1} \frac{du}{3x^2 + 2} = \int (x^2 + 2) \frac{3x^2 + 2x}{3x^2 + 2} t^{-1} du.$$\n\n27. **Simplify fraction:**\n$$\frac{3x^2 + 2x}{3x^2 + 2} = 1 + \frac{2x - 2}{3x^2 + 2}.$$\n\n28. **Integral becomes:**\n$$\int (x^2 + 2) \left(1 + \frac{2x - 2}{3x^2 + 2}\right) t^{-1} du = \int (x^2 + 2) t^{-1} du + \int (x^2 + 2) \frac{2x - 2}{3x^2 + 2} t^{-1} du.$$\n\n29. **Since $t = u^{1/3}$, $t^{-1} = u^{-1/3}$, and $du$ is differential of $u$, the integral is complicated to express purely in $u$.**\n\n30. **Conclusion:** The integral is nontrivial and does not simplify easily with elementary substitutions. It likely requires advanced techniques or numerical methods.\n\n**Final answer:** The integral $$\int \frac{(x^2 + 2)(3x^2 + 2x)}{\sqrt[3]{x^3 + 2x + 1}} \, dx$$ does not simplify to elementary functions by standard substitution methods and may require special functions or numerical integration.