Integral Evaluation D9D854

1. **State the problem:** We need to evaluate the definite integral $$\int_{1}^{10} \frac{2624}{x^{2} - x - 1} \, dx.$$ 2. **Rewrite the integral:** The integrand is $$\frac{2624}{x^{2} - x - 1}.$$ We will use partial fraction decomposition or substitution after factoring the denominator if possible. 3. **Factor the denominator:** The quadratic $$x^{2} - x - 1$$ has roots given by the quadratic formula: $$x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}.$$ So, $$x^{2} - x - 1 = (x - \alpha)(x - \beta)$$ where $$\alpha = \frac{1 + \sqrt{5}}{2}$$ and $$\beta = \frac{1 - \sqrt{5}}{2}.$$ 4. **Partial fraction decomposition:** We write $$\frac{2624}{(x - \alpha)(x - \beta)} = \frac{A}{x - \alpha} + \frac{B}{x - \beta}.$$ Multiply both sides by the denominator: $$2624 = A(x - \beta) + B(x - \alpha).$$ 5. **Find A and B:** Set $$x = \alpha$$ to find $$A$$: $$2624 = A(\alpha - \beta) + B(0) \Rightarrow A = \frac{2624}{\alpha - \beta}.$$ Set $$x = \beta$$ to find $$B$$: $$2624 = A(0) + B(\beta - \alpha) \Rightarrow B = \frac{2624}{\beta - \alpha} = -\frac{2624}{\alpha - \beta}.$$ 6. **Calculate $$\alpha - \beta$$:** $$\alpha - \beta = \frac{1 + \sqrt{5}}{2} - \frac{1 - \sqrt{5}}{2} = \frac{1 + \sqrt{5} - 1 + \sqrt{5}}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5}.$$ 7. **Substitute values for A and B:** $$A = \frac{2624}{\sqrt{5}}, \quad B = -\frac{2624}{\sqrt{5}}.$$ 8. **Rewrite the integral:** $$\int_{1}^{10} \frac{2624}{x^{2} - x - 1} \, dx = \int_{1}^{10} \left( \frac{A}{x - \alpha} + \frac{B}{x - \beta} \right) dx = A \int_{1}^{10} \frac{1}{x - \alpha} dx + B \int_{1}^{10} \frac{1}{x - \beta} dx.$$ 9. **Integrate:** $$\int \frac{1}{x - c} dx = \ln|x - c| + C.$$ So, $$= A [\ln|x - \alpha|]_{1}^{10} + B [\ln|x - \beta|]_{1}^{10} = A (\ln|10 - \alpha| - \ln|1 - \alpha|) + B (\ln|10 - \beta| - \ln|1 - \beta|).$$ 10. **Combine logarithms:** $$= A \ln \frac{|10 - \alpha|}{|1 - \alpha|} + B \ln \frac{|10 - \beta|}{|1 - \beta|}.$$ 11. **Substitute A and B:** $$= \frac{2624}{\sqrt{5}} \ln \frac{|10 - \alpha|}{|1 - \alpha|} - \frac{2624}{\sqrt{5}} \ln \frac{|10 - \beta|}{|1 - \beta|} = \frac{2624}{\sqrt{5}} \left( \ln \frac{|10 - \alpha|}{|1 - \alpha|} - \ln \frac{|10 - \beta|}{|1 - \beta|} \right).$$ 12. **Use logarithm properties:** $$= \frac{2624}{\sqrt{5}} \ln \left( \frac{|10 - \alpha|}{|1 - \alpha|} \cdot \frac{|1 - \beta|}{|10 - \beta|} \right).$$ 13. **Calculate numerical values:** $$\alpha = \frac{1 + \sqrt{5}}{2} \approx 1.618, \quad \beta = \frac{1 - \sqrt{5}}{2} \approx -0.618.$$ Calculate each term: $$|10 - \alpha| = 10 - 1.618 = 8.382,$$ $$|1 - \alpha| = |1 - 1.618| = 0.618,$$ $$|10 - \beta| = 10 - (-0.618) = 10.618,$$ $$|1 - \beta| = 1 - (-0.618) = 1.618.$$ 14. **Substitute:** $$\frac{|10 - \alpha|}{|1 - \alpha|} \cdot \frac{|1 - \beta|}{|10 - \beta|} = \frac{8.382}{0.618} \cdot \frac{1.618}{10.618} = 13.56 \times 0.1524 = 2.066.$$ 15. **Final answer:** $$\int_{1}^{10} \frac{2624}{x^{2} - x - 1} \, dx = \frac{2624}{\sqrt{5}} \ln(2.066) \approx \frac{2624}{2.236} \times 0.726 = 1173.3.$$ **Therefore, the value of the integral is approximately** $$1173.3.$$