1. **State the problem:** We need to solve the differential equation $$\frac{dy}{dx} = \frac{2x + 4y + 3}{2y + x + 1}.$$\n\n2. **Analyze the equation:** This is a first-order differential equation given as a ratio of linear expressions in $x$ and $y$. We want to find $y$ as a function of $x$.\n\n3. **Rewrite the equation:** Multiply both sides by the denominator to get\n$$ (2y + x + 1) dy = (2x + 4y + 3) dx. $$\n\n4. **Rewrite in differential form:**\n$$ (2y + x + 1) dy - (2x + 4y + 3) dx = 0. $$\n\n5. **Check if the differential form is exact:** Let \(M = 2y + x + 1\) (coefficient of $dy$) and \(N = -(2x + 4y + 3)\) (coefficient of $dx$).\nCalculate partial derivatives:\n$$ \frac{\partial M}{\partial x} = 1, \quad \frac{\partial N}{\partial y} = -4. $$\nSince \(\frac{\partial M}{\partial x} \neq \frac{\partial N}{\partial y}\), the equation is not exact.\n\n6. **Try to find an integrating factor:** Since $M$ and $N$ are linear in $x$ and $y$, try an integrating factor depending on $x$ or $y$.\nCheck if \(\frac{\partial}{\partial y} \left( \frac{M}{N} \right)\) or \(\frac{\partial}{\partial x} \left( \frac{N}{M} \right)\) is a function of one variable. Alternatively, try substitution.\n\n7. **Try substitution:** Let\n$$ u = 2y + x + 1. $$\nThen\n$$ du = 2 dy + dx. $$\nExpress $dy$ and $dx$ in terms of $du$ and $dx$ or $dy$.\n\n8. **Rewrite the original equation in terms of $u$:**\nNote that numerator is $2x + 4y + 3$. Express $4y$ as $2(2y)$ and use $u = 2y + x + 1$ to rewrite terms.\n\n9. **Express numerator:**\n$$ 2x + 4y + 3 = 2x + 2(2y) + 3 = 2x + 2(u - x - 1) + 3 = 2x + 2u - 2x - 2 + 3 = 2u + 1. $$\n\n10. **Rewrite denominator:**\n$$ 2y + x + 1 = u. $$\n\n11. **Rewrite differential equation:**\n$$ \frac{dy}{dx} = \frac{2u + 1}{u}. $$\n\n12. **Express $dy/dx$ in terms of $u$ and $x$:**\nFrom $u = 2y + x + 1$, differentiate both sides with respect to $x$:\n$$ \frac{du}{dx} = 2 \frac{dy}{dx} + 1. $$\n\n13. **Substitute $dy/dx$ from step 11:**\n$$ \frac{du}{dx} = 2 \cdot \frac{2u + 1}{u} + 1 = \frac{4u + 2}{u} + 1 = \frac{4u + 2 + u}{u} = \frac{5u + 2}{u}. $$\n\n14. **Rewrite as:**\n$$ \frac{du}{dx} = 5 + \frac{2}{u}. $$\n\n15. **Separate variables:**\n$$ \frac{du}{5 + \frac{2}{u}} = dx. $$\nMultiply numerator and denominator by $u$ to simplify denominator:\n$$ \frac{du}{5 + \frac{2}{u}} = \frac{du}{\frac{5u + 2}{u}} = \frac{u}{5u + 2} du = dx. $$\n\n16. **Integrate both sides:**\n$$ \int \frac{u}{5u + 2} du = \int dx. $$\n\n17. **Simplify the integral on the left:**\nRewrite numerator:\n$$ u = \frac{1}{5}(5u + 2) - \frac{2}{5}. $$\nSo\n$$ \int \frac{u}{5u + 2} du = \int \frac{\frac{1}{5}(5u + 2) - \frac{2}{5}}{5u + 2} du = \int \left( \frac{1}{5} - \frac{2}{5(5u + 2)} \right) du. $$\n\n18. **Integrate term by term:**\n$$ \int \frac{1}{5} du - \frac{2}{5} \int \frac{1}{5u + 2} du = \frac{u}{5} - \frac{2}{5} \cdot \frac{1}{5} \ln|5u + 2| + C = \frac{u}{5} - \frac{2}{25} \ln|5u + 2| + C. $$\n\n19. **Integrate right side:**\n$$ \int dx = x + C'. $$\n\n20. **Combine constants and write implicit solution:**\n$$ \frac{u}{5} - \frac{2}{25} \ln|5u + 2| = x + C''. $$\n\n21. **Recall substitution $u = 2y + x + 1$:**\n$$ \frac{2y + x + 1}{5} - \frac{2}{25} \ln|5(2y + x + 1) + 2| = x + C. $$\n\n22. **Final implicit solution:**\n$$ \boxed{\frac{2y + x + 1}{5} - \frac{2}{25} \ln|10y + 5x + 7| = x + C}. $$\n\nThis expresses $y$ implicitly as a function of $x$.