1. **State the problem:** We want to evaluate the integral $$\int_1^{\infty} (4 + 2x + 6x^2) e^{-(5 + 4x + x^2 + 2x^3)} \, dx.$$\n\n2. **Analyze the integrand:** The integrand is a product of a polynomial $(4 + 2x + 6x^2)$ and an exponential function with a complicated exponent $-(5 + 4x + x^2 + 2x^3)$.\n\n3. **Consider substitution:** Let $$f(x) = 5 + 4x + x^2 + 2x^3.$$ Then the integrand is $(4 + 2x + 6x^2) e^{-f(x)}$.\n\n4. **Check if the polynomial is related to the derivative of $f(x)$:** Compute $$f'(x) = 4 + 2x + 6x^2,$$ which exactly matches the polynomial factor in the integrand.\n\n5. **Rewrite the integral using substitution:** Since $f'(x) = 4 + 2x + 6x^2$, we have $$ (4 + 2x + 6x^2) e^{-f(x)} = f'(x) e^{-f(x)}.$$\n\n6. **Use substitution $u = f(x)$:** Then $$du = f'(x) dx,$$ so $$f'(x) dx = du.$$ The integral becomes $$\int_1^{\infty} f'(x) e^{-f(x)} dx = \int_{f(1)}^{f(\infty)} e^{-u} du.$$\n\n7. **Evaluate the limits:** Calculate $$f(1) = 5 + 4(1) + 1^2 + 2(1)^3 = 5 + 4 + 1 + 2 = 12.$$ As $x \to \infty$, the dominant term in $f(x)$ is $2x^3$, so $$f(\infty) = \infty.$$\n\n8. **Evaluate the integral in $u$:** $$\int_{12}^{\infty} e^{-u} du = \left[-e^{-u}\right]_{12}^{\infty} = 0 - (-e^{-12}) = e^{-12}.$$\n\n**Final answer:** $$\boxed{e^{-12}}.$$