1. **State the problem:** We need to integrate $$\int \frac{x^3 + 1}{x^2 + 7x + 12} \, dx$$ by using polynomial long division. 2. **Perform long division:** Divide the polynomial numerator $$x^3 + 1$$ by the denominator $$x^2 + 7x + 12$$. - Divide $$x^3$$ by $$x^2$$ to get $$x$$. - Multiply $$x$$ by $$x^2 + 7x + 12$$ to get $$x^3 + 7x^2 + 12x$$. - Subtract this from the numerator: $$ (x^3 + 1) - (x^3 + 7x^2 + 12x) = -7x^2 - 12x + 1 $$. 3. **Continue dividing:** Divide $$-7x^2$$ by $$x^2$$ to get $$-7$$. - Multiply $$-7$$ by $$x^2 + 7x + 12$$ to get $$-7x^2 - 49x - 84$$. - Subtract from previous remainder: $$ (-7x^2 - 12x + 1) - (-7x^2 - 49x - 84) = 37x + 85 $$. 4. So the quotient is $$x - 7$$ and the remainder is $$37x + 85$$. Rewrite the integrand: $$\frac{x^3 + 1}{x^2 + 7x + 12} = x - 7 + \frac{37x + 85}{x^2 + 7x + 12}$$. 5. **Factor the denominator:** $$x^2 + 7x + 12 = (x + 3)(x + 4)$$. 6. **Decompose the fraction:** Assume $$\frac{37x + 85}{(x + 3)(x + 4)} = \frac{A}{x + 3} + \frac{B}{x + 4}$$. Multiply both sides by $$ (x + 3)(x + 4) $$: $$37x + 85 = A(x + 4) + B(x + 3) = (A + B)x + (4A + 3B)$$. Equate coefficients: - Coefficient of $$x$$: $$37 = A + B$$ - Constant term: $$85 = 4A + 3B$$ 7. **Solve the system:** From $$37 = A + B$$, we get $$B = 37 - A$$. Substitute into the second equation: $$85 = 4A + 3(37 - A) = 4A + 111 - 3A = A + 111$$. So $$A = 85 - 111 = -26$$. Then $$B = 37 - (-26) = 63$$. 8. **Rewrite the integral:** $$\int \left(x - 7 + \frac{-26}{x + 3} + \frac{63}{x + 4}\right) dx$$ 9. **Integrate term by term:** - $$\int x \, dx = \frac{x^2}{2}$$ - $$\int -7 \, dx = -7x$$ - $$\int \frac{-26}{x + 3} \, dx = -26 \ln|x + 3|$$ - $$\int \frac{63}{x + 4} \, dx = 63 \ln|x + 4|$$ 10. **Combine results:** $$\frac{x^2}{2} - 7x - 26 \ln|x + 3| + 63 \ln|x + 4| + C$$ **Final answer:** $$\int \frac{x^3 + 1}{x^2 + 7x + 12} \, dx = \frac{x^2}{2} - 7x - 26 \ln|x + 3| + 63 \ln|x + 4| + C$$