1. We are asked to evaluate the definite integral $$\int_0^1 \frac{1 + 6t}{1 + 2t} \, dt$$. 2. To solve this, we use algebraic manipulation to simplify the integrand. Notice that the numerator and denominator are linear expressions. 3. Perform polynomial division or rewrite the fraction: $$\frac{1 + 6t}{1 + 2t} = A + \frac{B}{1 + 2t}$$ for constants $A$ and $B$. 4. Multiply both sides by $1 + 2t$: $$1 + 6t = A(1 + 2t) + B$$ $$1 + 6t = A + 2At + B$$ 5. Group terms: $$1 + 6t = (A + B) + 2At$$ 6. Equate coefficients of like terms: For $t$: $6 = 2A \implies A = 3$ For constants: $1 = A + B = 3 + B \implies B = 1 - 3 = -2$ 7. So the integrand becomes: $$\frac{1 + 6t}{1 + 2t} = 3 - \frac{2}{1 + 2t}$$ 8. Rewrite the integral: $$\int_0^1 \frac{1 + 6t}{1 + 2t} \, dt = \int_0^1 \left(3 - \frac{2}{1 + 2t}\right) dt = \int_0^1 3 \, dt - 2 \int_0^1 \frac{1}{1 + 2t} \, dt$$ 9. Integrate each term separately: $$\int_0^1 3 \, dt = 3t \Big|_0^1 = 3(1) - 3(0) = 3$$ 10. For the second integral, use substitution: Let $u = 1 + 2t$, then $du = 2 dt$, so $dt = \frac{du}{2}$. When $t=0$, $u=1$; when $t=1$, $u=3$. 11. Substitute: $$\int_0^1 \frac{1}{1 + 2t} \, dt = \int_1^3 \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \int_1^3 \frac{1}{u} \, du = \frac{1}{2} \ln|u| \Big|_1^3 = \frac{1}{2} (\ln 3 - \ln 1) = \frac{1}{2} \ln 3$$ 12. Multiply by -2: $$-2 \times \frac{1}{2} \ln 3 = -\ln 3$$ 13. Combine results: $$3 - \ln 3$$ 14. Therefore, the value of the integral is: $$\boxed{3 - \ln 3}$$