1. **State the problem:** Evaluate the definite integral $$\int_1^4 \frac{4x+1}{2x^2+x} \, dx$$. 2. **Simplify the integrand:** Factor the denominator: $$2x^2 + x = x(2x+1)$$ So the integral becomes: $$\int_1^4 \frac{4x+1}{x(2x+1)} \, dx$$. 3. **Use partial fraction decomposition:** Assume $$\frac{4x+1}{x(2x+1)} = \frac{A}{x} + \frac{B}{2x+1}$$ Multiply both sides by the denominator: $$4x+1 = A(2x+1) + Bx$$ Expand: $$4x+1 = 2Ax + A + Bx = (2A + B)x + A$$ Equate coefficients: $$2A + B = 4$$ $$A = 1$$ From $A=1$, substitute into first equation: $$2(1) + B = 4 \implies B = 2$$ 4. **Rewrite the integral:** $$\int_1^4 \left( \frac{1}{x} + \frac{2}{2x+1} \right) dx$$ 5. **Integrate term-by-term:** $$\int_1^4 \frac{1}{x} dx + \int_1^4 \frac{2}{2x+1} dx$$ 6. **First integral:** $$\int_1^4 \frac{1}{x} dx = \left[ \ln|x| \right]_1^4 = \ln 4 - \ln 1 = \ln 4$$ 7. **Second integral:** Use substitution $u = 2x+1$, so $du = 2 dx$, or $dx = \frac{du}{2}$. Change limits: When $x=1$, $u=3$. When $x=4$, $u=9$. Rewrite integral: $$\int_3^9 \frac{2}{u} \cdot \frac{du}{2} = \int_3^9 \frac{1}{u} du = \left[ \ln|u| \right]_3^9 = \ln 9 - \ln 3 = \ln \frac{9}{3} = \ln 3$$ 8. **Add results:** $$\ln 4 + \ln 3 = \ln (4 \times 3) = \ln 12$$ **Final answer:** $$\boxed{\ln 12}$$