1. **Problem a:** Calculate the definite integral $$\int_0^1 \ln(x+2) \, dx$$ using integration by parts. 2. **Formula for integration by parts:** $$\int u \, dv = uv - \int v \, du$$ 3. **Choose:** - Let $$u = \ln(x+2)$$ so $$du = \frac{1}{x+2} dx$$ - Let $$dv = dx$$ so $$v = x$$ 4. **Apply integration by parts:** $$\int_0^1 \ln(x+2) \, dx = \left. x \ln(x+2) \right|_0^1 - \int_0^1 \frac{x}{x+2} dx$$ 5. **Evaluate the boundary term:** $$1 \cdot \ln(3) - 0 \cdot \ln(2) = \ln(3)$$ 6. **Simplify the integral:** $$\int_0^1 \frac{x}{x+2} dx = \int_0^1 \frac{x+2-2}{x+2} dx = \int_0^1 \left(1 - \frac{2}{x+2}\right) dx = \int_0^1 1 \, dx - 2 \int_0^1 \frac{1}{x+2} dx$$ 7. **Calculate each integral:** - $$\int_0^1 1 \, dx = 1$$ - $$\int_0^1 \frac{1}{x+2} dx = \left. \ln|x+2| \right|_0^1 = \ln(3) - \ln(2) = \ln\left(\frac{3}{2}\right)$$ 8. **Combine:** $$\int_0^1 \frac{x}{x+2} dx = 1 - 2 \ln\left(\frac{3}{2}\right)$$ 9. **Final answer for a:** $$\int_0^1 \ln(x+2) \, dx = \ln(3) - \left(1 - 2 \ln\left(\frac{3}{2}\right)\right) = \ln(3) - 1 + 2 \ln\left(\frac{3}{2}\right)$$ --- 10. **Problem b:** Calculate the definite integral $$\int_1^5 \frac{x+7}{x^3 + 2x^2} dx$$ using partial fractions. 11. **Factor denominator:** $$x^3 + 2x^2 = x^2(x+2)$$ 12. **Set up partial fractions:** $$\frac{x+7}{x^2(x+2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+2}$$ 13. **Multiply both sides by denominator:** $$x+7 = A x (x+2) + B (x+2) + C x^2$$ 14. **Expand right side:** $$A x^2 + 2 A x + B x + 2 B + C x^2 = (A + C) x^2 + (2A + B) x + 2 B$$ 15. **Match coefficients:** - Coefficient of $$x^2$$: $$0 = A + C$$ - Coefficient of $$x$$: $$1 = 2A + B$$ - Constant term: $$7 = 2B$$ 16. **Solve system:** - From constant: $$B = \frac{7}{2}$$ - From $$x^2$$: $$C = -A$$ - From $$x$$: $$1 = 2A + \frac{7}{2} \Rightarrow 2A = 1 - \frac{7}{2} = -\frac{5}{2} \Rightarrow A = -\frac{5}{4}$$ - Then $$C = -A = \frac{5}{4}$$ 17. **Rewrite integral:** $$\int_1^5 \left( \frac{-\frac{5}{4}}{x} + \frac{\frac{7}{2}}{x^2} + \frac{\frac{5}{4}}{x+2} \right) dx = \int_1^5 \left(-\frac{5}{4x} + \frac{7}{2x^2} + \frac{5}{4(x+2)}\right) dx$$ 18. **Integrate term by term:** - $$\int \frac{1}{x} dx = \ln|x|$$ - $$\int x^{-2} dx = -x^{-1}$$ - $$\int \frac{1}{x+2} dx = \ln|x+2|$$ 19. **Evaluate definite integral:** $$= \left[-\frac{5}{4} \ln|x| - \frac{7}{2x} + \frac{5}{4} \ln|x+2| \right]_1^5$$ 20. **Calculate at bounds:** - At $$x=5$$: $$-\frac{5}{4} \ln 5 - \frac{7}{10} + \frac{5}{4} \ln 7$$ - At $$x=1$$: $$-\frac{5}{4} \ln 1 - \frac{7}{2} + \frac{5}{4} \ln 3 = 0 - \frac{7}{2} + \frac{5}{4} \ln 3$$ 21. **Subtract:** $$\left(-\frac{5}{4} \ln 5 - \frac{7}{10} + \frac{5}{4} \ln 7\right) - \left(0 - \frac{7}{2} + \frac{5}{4} \ln 3\right) = -\frac{5}{4} \ln 5 - \frac{7}{10} + \frac{5}{4} \ln 7 + \frac{7}{2} - \frac{5}{4} \ln 3$$ 22. **Simplify constants:** $$-\frac{7}{10} + \frac{7}{2} = \frac{-7 + 35}{10} = \frac{28}{10} = \frac{14}{5}$$ 23. **Final answer for b:** $$\frac{14}{5} + \frac{5}{4} \ln\left(\frac{7}{15}\right)$$ --- **Summary:** - a) $$\int_0^1 \ln(x+2) dx = \ln(3) - 1 + 2 \ln\left(\frac{3}{2}\right)$$ - b) $$\int_1^5 \frac{x+7}{x^3 + 2x^2} dx = \frac{14}{5} + \frac{5}{4} \ln\left(\frac{7}{15}\right)$$