1. **State the problem:** We want to evaluate the integral $$\int_0^6 \frac{x^2 + 8x - 3}{x + 2} \, dx$$ and express the answer in the form $a + b \ln 2$ where $a$ and $b$ are integers. 2. **Rewrite the integrand using polynomial division:** Divide $x^2 + 8x - 3$ by $x + 2$ to express it as $$\frac{x^2 + 8x - 3}{x + 2} = Ax + B + \frac{C}{x + 2}$$ where $A$, $B$, and $C$ are constants. 3. **Perform polynomial division:** Divide $x^2 + 8x - 3$ by $x + 2$: $$x^2 + 8x - 3 \div (x + 2) = x + 6 + \frac{9}{x + 2}$$ Explanation: - $x \times (x + 2) = x^2 + 2x$ - Subtract: $(x^2 + 8x - 3) - (x^2 + 2x) = 6x - 3$ - $6 \times (x + 2) = 6x + 12$ - Subtract: $(6x - 3) - (6x + 12) = -15$ Correction: The remainder is $-15$, so the division is actually: $$x^2 + 8x - 3 = (x + 2)(x + 6) - 15$$ So, $$\frac{x^2 + 8x - 3}{x + 2} = x + 6 - \frac{15}{x + 2}$$ 4. **Rewrite the integral:** $$\int_0^6 \frac{x^2 + 8x - 3}{x + 2} \, dx = \int_0^6 \left(x + 6 - \frac{15}{x + 2}\right) dx$$ 5. **Integrate term-by-term:** $$\int_0^6 x \, dx = \left[\frac{x^2}{2}\right]_0^6 = \frac{36}{2} = 18$$ $$\int_0^6 6 \, dx = [6x]_0^6 = 36$$ $$\int_0^6 \frac{15}{x + 2} \, dx = 15 \int_0^6 \frac{1}{x + 2} \, dx = 15 [\ln|x + 2|]_0^6 = 15 (\ln 8 - \ln 2) = 15 \ln 4$$ 6. **Combine the results:** $$\int_0^6 \frac{x^2 + 8x - 3}{x + 2} \, dx = 18 + 36 - 15 \ln 4 = 54 - 15 \ln 4$$ 7. **Express $\ln 4$ in terms of $\ln 2$:** $$\ln 4 = \ln (2^2) = 2 \ln 2$$ So, $$54 - 15 \ln 4 = 54 - 15 \times 2 \ln 2 = 54 - 30 \ln 2$$ 8. **Final answer:** $$\boxed{54 - 30 \ln 2}$$ where $a = 54$ and $b = -30$.