1. **State the problem:** Solve the differential equation $$(x^3+x^2+x+1)\frac{dy}{dx} = 2x^2 + x$$ with the initial condition $x=0$, $y=1$. 2. **Rewrite the equation:** Isolate $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{2x^2 + x}{x^3 + x^2 + x + 1}$$ 3. **Simplify the denominator:** Factor the denominator: $$x^3 + x^2 + x + 1 = (x^2 + 1)(x + 1)$$ 4. **Rewrite the derivative:** $$\frac{dy}{dx} = \frac{2x^2 + x}{(x^2 + 1)(x + 1)}$$ 5. **Use partial fraction decomposition:** Express $$\frac{2x^2 + x}{(x^2 + 1)(x + 1)} = \frac{Ax + B}{x^2 + 1} + \frac{C}{x + 1}$$ Multiply both sides by $(x^2 + 1)(x + 1)$: $$2x^2 + x = (Ax + B)(x + 1) + C(x^2 + 1)$$ 6. **Expand and collect terms:** $$(Ax + B)(x + 1) = Ax^2 + Ax + Bx + B = Ax^2 + (A + B)x + B$$ So, $$2x^2 + x = Ax^2 + (A + B)x + B + Cx^2 + C$$ $$2x^2 + x = (A + C)x^2 + (A + B)x + (B + C)$$ 7. **Equate coefficients:** - For $x^2$: $2 = A + C$ - For $x$: $1 = A + B$ - For constant: $0 = B + C$ 8. **Solve the system:** From $0 = B + C$, we get $B = -C$. Substitute into $1 = A + B$: $1 = A - C$. Substitute into $2 = A + C$. Add the last two equations: $$(1 = A - C) + (2 = A + C) \Rightarrow 3 = 2A \Rightarrow A = \frac{3}{2}$$ From $2 = A + C$, $$2 = \frac{3}{2} + C \Rightarrow C = 2 - \frac{3}{2} = \frac{1}{2}$$ From $B = -C$, $$B = -\frac{1}{2}$$ 9. **Rewrite the derivative:** $$\frac{dy}{dx} = \frac{\frac{3}{2}x - \frac{1}{2}}{x^2 + 1} + \frac{\frac{1}{2}}{x + 1}$$ 10. **Integrate both sides:** $$y = \int \left( \frac{\frac{3}{2}x - \frac{1}{2}}{x^2 + 1} + \frac{\frac{1}{2}}{x + 1} \right) dx + C$$ Split the integral: $$y = \frac{3}{2} \int \frac{x}{x^2 + 1} dx - \frac{1}{2} \int \frac{1}{x^2 + 1} dx + \frac{1}{2} \int \frac{1}{x + 1} dx + C$$ 11. **Calculate each integral:** - $$\int \frac{x}{x^2 + 1} dx$$ Let $u = x^2 + 1$, $du = 2x dx$, so $$\int \frac{x}{x^2 + 1} dx = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|x^2 + 1|$$ - $$\int \frac{1}{x^2 + 1} dx = \arctan x$$ - $$\int \frac{1}{x + 1} dx = \ln|x + 1|$$ 12. **Substitute back:** $$y = \frac{3}{2} \cdot \frac{1}{2} \ln|x^2 + 1| - \frac{1}{2} \arctan x + \frac{1}{2} \ln|x + 1| + C$$ $$y = \frac{3}{4} \ln(x^2 + 1) - \frac{1}{2} \arctan x + \frac{1}{2} \ln|x + 1| + C$$ 13. **Apply initial condition $y(0) = 1$:** $$1 = \frac{3}{4} \ln(0^2 + 1) - \frac{1}{2} \arctan 0 + \frac{1}{2} \ln|0 + 1| + C$$ $$1 = \frac{3}{4} \cdot 0 - 0 + \frac{1}{2} \cdot 0 + C = C$$ So, $$C = 1$$. 14. **Final solution:** $$\boxed{y = \frac{3}{4} \ln(x^2 + 1) - \frac{1}{2} \arctan x + \frac{1}{2} \ln|x + 1| + 1}$$