1. **Problem:** Find the exact value of $$\int_1^3 x^2 \ln(3x) \, dx$$ in the form $$a \ln b + c$$ where $$a,c$$ are rational and $$b$$ is an integer. 2. **Formula and rules:** Use integration by parts: $$\int u \, dv = uv - \int v \, du$$. 3. **Step 1:** Let $$u = \ln(3x)$$ and $$dv = x^2 dx$$. 4. Then $$du = \frac{1}{x} dx$$ and $$v = \frac{x^3}{3}$$. 5. **Step 2:** Apply integration by parts: $$\int_1^3 x^2 \ln(3x) dx = \left. \frac{x^3}{3} \ln(3x) \right|_1^3 - \int_1^3 \frac{x^3}{3} \cdot \frac{1}{x} dx = \left. \frac{x^3}{3} \ln(3x) \right|_1^3 - \frac{1}{3} \int_1^3 x^2 dx$$ 6. **Step 3:** Evaluate the integral: $$\int_1^3 x^2 dx = \left. \frac{x^3}{3} \right|_1^3 = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$$ 7. **Step 4:** Evaluate the boundary term: $$\left. \frac{x^3}{3} \ln(3x) \right|_1^3 = \frac{27}{3} \ln(9) - \frac{1}{3} \ln(3) = 9 \ln(9) - \frac{1}{3} \ln(3)$$ 8. **Step 5:** Substitute back: $$9 \ln(9) - \frac{1}{3} \ln(3) - \frac{1}{3} \cdot \frac{26}{3} = 9 \ln(9) - \frac{1}{3} \ln(3) - \frac{26}{9}$$ 9. **Step 6:** Simplify logarithms: $$9 \ln(9) = 9 \ln(3^2) = 18 \ln(3)$$ 10. So the answer is: $$18 \ln(3) - \frac{1}{3} \ln(3) - \frac{26}{9} = \left(18 - \frac{1}{3}\right) \ln(3) - \frac{26}{9} = \frac{53}{3} \ln(3) - \frac{26}{9}$$ --- 11. **Problem 2(a):** Given $$f(x) = \frac{2 e^{2x}}{e^{2x} - 3 e^x + 2}$$, find $$f'(x)$$ and the stationary points. 12. **Step 1:** Let $$g(x) = e^{2x} - 3 e^x + 2$$, then $$f(x) = \frac{2 e^{2x}}{g(x)}$$. 13. Use quotient rule: $$f'(x) = \frac{(2 e^{2x})' g(x) - 2 e^{2x} g'(x)}{g(x)^2}$$ 14. Compute derivatives: $$(2 e^{2x})' = 4 e^{2x}$$ $$(g(x))' = 2 e^{2x} - 3 e^x$$ 15. Substitute: $$f'(x) = \frac{4 e^{2x} (e^{2x} - 3 e^x + 2) - 2 e^{2x} (2 e^{2x} - 3 e^x)}{(e^{2x} - 3 e^x + 2)^2}$$ 16. Simplify numerator: $$4 e^{4x} - 12 e^{3x} + 8 e^{2x} - 4 e^{4x} + 6 e^{3x} = (-6 e^{3x} + 8 e^{2x})$$ 17. So: $$f'(x) = \frac{-6 e^{3x} + 8 e^{2x}}{(e^{2x} - 3 e^x + 2)^2} = \frac{2 e^{2x} (-3 e^x + 4)}{(e^{2x} - 3 e^x + 2)^2}$$ 18. **Step 2:** Stationary points where $$f'(x) = 0$$, numerator zero: $$2 e^{2x} (-3 e^x + 4) = 0 \implies -3 e^x + 4 = 0 \implies e^x = \frac{4}{3}$$ 19. So stationary point at $$x = \ln \frac{4}{3}$$. 20. Find $$y$$ coordinate: $$f\left(\ln \frac{4}{3}\right) = \frac{2 e^{2 \ln \frac{4}{3}}}{e^{2 \ln \frac{4}{3}} - 3 e^{\ln \frac{4}{3}} + 2} = \frac{2 \left(\frac{4}{3}\right)^2}{\left(\frac{4}{3}\right)^2 - 3 \cdot \frac{4}{3} + 2} = \frac{2 \cdot \frac{16}{9}}{\frac{16}{9} - 4 + 2} = \frac{\frac{32}{9}}{\frac{16}{9} - 2} = \frac{\frac{32}{9}}{\frac{16}{9} - \frac{18}{9}} = \frac{\frac{32}{9}}{-\frac{2}{9}} = -16$$ 21. Stationary point: $$\left(\ln \frac{4}{3}, -16\right)$$. --- 22. **Problem 2(b):** Find $$\int_{\ln 3}^{\ln 5} f(x) dx$$ using substitution $$u = e^x$$ and partial fractions. 23. Substitute $$u = e^x$$, so $$du = e^x dx = u dx \implies dx = \frac{du}{u}$$. 24. Change limits: When $$x = \ln 3$$, $$u = 3$$. When $$x = \ln 5$$, $$u = 5$$. 25. Rewrite integral: $$\int_{\ln 3}^{\ln 5} \frac{2 e^{2x}}{e^{2x} - 3 e^x + 2} dx = \int_3^5 \frac{2 u^2}{u^2 - 3 u + 2} \cdot \frac{1}{u} du = \int_3^5 \frac{2 u}{u^2 - 3 u + 2} du$$ 26. Factor denominator: $$u^2 - 3 u + 2 = (u - 1)(u - 2)$$ 27. Partial fractions: $$\frac{2 u}{(u - 1)(u - 2)} = \frac{A}{u - 1} + \frac{B}{u - 2}$$ 28. Multiply both sides by denominator: $$2 u = A (u - 2) + B (u - 1) = (A + B) u - 2 A - B$$ 29. Equate coefficients: $$A + B = 2$$ $$-2 A - B = 0$$ 30. Solve system: From second: $$B = -2 A$$ Substitute into first: $$A - 2 A = 2 \implies -A = 2 \implies A = -2$$ Then $$B = -2 (-2) = 4$$ 31. So: $$\frac{2 u}{(u - 1)(u - 2)} = \frac{-2}{u - 1} + \frac{4}{u - 2}$$ 32. Integral becomes: $$\int_3^5 \left( \frac{-2}{u - 1} + \frac{4}{u - 2} \right) du = -2 \int_3^5 \frac{1}{u - 1} du + 4 \int_3^5 \frac{1}{u - 2} du$$ 33. Integrate: $$-2 [\ln|u - 1|]_3^5 + 4 [\ln|u - 2|]_3^5 = -2 (\ln 4 - \ln 2) + 4 (\ln 3 - \ln 1)$$ 34. Simplify: $$-2 \ln 2 + 4 \ln 3 = \ln 3^4 - \ln 2^2 = \ln \frac{81}{4}$$ 35. **Answer:** $$\int_{\ln 3}^{\ln 5} f(x) dx = \ln \frac{81}{4}$$. --- 36. **Problem 3(a):** Find $$a$$ where $$y = 2 \sin x \sqrt{2} + \cos x$$ has minimum point $$M$$ at $$x = a$$ between $$\pi$$ and $$2 \pi$$. 37. **Step 1:** Find $$y'$$: $$y' = 2 \sqrt{2} \cos x - \sin x$$ 38. Set $$y' = 0$$ for stationary points: $$2 \sqrt{2} \cos x - \sin x = 0 \implies \tan x = 2 \sqrt{2}$$ 39. Find $$x$$ in $$[\pi, 2 \pi]$$: $$x = \arctan(2 \sqrt{2}) + \pi \approx 1.895 + 3.142 = 5.037$$ 40. Rounded to 2 decimals: $$a = 5.04$$ --- 41. **Problem 3(b):** Find exact area of shaded region $$R$$ using substitution $$u = 2 + \cos x$$. 42. Area under curve from $$a$$ to $$2 \pi$$: $$A = \int_a^{2 \pi} y \, dx = \int_a^{2 \pi} (2 \sin x \sqrt{2} + \cos x) dx$$ 43. Substitute $$u = 2 + \cos x$$, so $$du = -\sin x dx$$. 44. Rewrite integral: $$\int_a^{2 \pi} 2 \sqrt{2} \sin x + \cos x \, dx = 2 \sqrt{2} \int_a^{2 \pi} \sin x \, dx + \int_a^{2 \pi} \cos x \, dx$$ 45. Use substitution for $$\sin x$$ term: $$\int_a^{2 \pi} \sin x \, dx = - \int_{u(a)}^{u(2 \pi)} du = - (u(2 \pi) - u(a)) = u(a) - u(2 \pi)$$ 46. Evaluate $$u$$ at limits: $$u(2 \pi) = 2 + \cos 2 \pi = 2 + 1 = 3$$ $$u(a) = 2 + \cos a$$ 47. So: $$2 \sqrt{2} (u(a) - 3) + [\sin x]_a^{2 \pi} = 2 \sqrt{2} (2 + \cos a - 3) + (\sin 2 \pi - \sin a) = 2 \sqrt{2} (\cos a - 1) - \sin a$$ 48. Use $$\tan a = 2 \sqrt{2}$$ and $$\sin a = \frac{\tan a}{\sqrt{1 + \tan^2 a}}$$, $$\cos a = \frac{1}{\sqrt{1 + \tan^2 a}}$$. 49. Calculate: $$\tan a = 2 \sqrt{2}$$ $$1 + \tan^2 a = 1 + 8 = 9$$ $$\sin a = \frac{2 \sqrt{2}}{3}$$ $$\cos a = \frac{1}{3}$$ 50. Substitute: $$2 \sqrt{2} \left( \frac{1}{3} - 1 \right) - \frac{2 \sqrt{2}}{3} = 2 \sqrt{2} \left( -\frac{2}{3} \right) - \frac{2 \sqrt{2}}{3} = -\frac{4 \sqrt{2}}{3} - \frac{2 \sqrt{2}}{3} = -2 \sqrt{2}$$ 51. Area is positive, so: $$A = 2 \sqrt{2}$$ --- 52. **Problem 4(a):** Express $$f(x) = \frac{-7 x^2 + 2 x - 6}{(1 + x)(4 + x^2)}$$ in partial fractions. 53. Assume: $$\frac{-7 x^2 + 2 x - 6}{(1 + x)(4 + x^2)} = \frac{A}{1 + x} + \frac{B x + C}{4 + x^2}$$ 54. Multiply both sides by denominator: $$-7 x^2 + 2 x - 6 = A (4 + x^2) + (B x + C)(1 + x) = 4 A + A x^2 + B x + B x^2 + C + C x$$ 55. Group terms: $$-7 x^2 + 2 x - 6 = (A + B) x^2 + (B + C) x + (4 A + C)$$ 56. Equate coefficients: $$x^2: A + B = -7$$ $$x: B + C = 2$$ $$constant: 4 A + C = -6$$ 57. Solve system: From first: $$B = -7 - A$$ From second: $$C = 2 - B = 2 - (-7 - A) = 9 + A$$ Substitute into third: $$4 A + 9 + A = -6 \implies 5 A = -15 \implies A = -3$$ 58. Then: $$B = -7 - (-3) = -4$$ $$C = 9 + (-3) = 6$$ 59. So partial fractions: $$f(x) = \frac{-3}{1 + x} + \frac{-4 x + 6}{4 + x^2}$$ --- 60. **Problem 4(b):** Find $$\int_0^2 f(x) dx$$ in form $$a \pi - \ln b$$. 61. Integral: $$\int_0^2 \left( \frac{-3}{1 + x} + \frac{-4 x + 6}{4 + x^2} \right) dx = -3 \int_0^2 \frac{1}{1 + x} dx + \int_0^2 \frac{-4 x}{4 + x^2} dx + \int_0^2 \frac{6}{4 + x^2} dx$$ 62. Evaluate each: $$-3 \int_0^2 \frac{1}{1 + x} dx = -3 [\ln|1 + x|]_0^2 = -3 (\ln 3 - \ln 1) = -3 \ln 3$$ 63. For $$\int_0^2 \frac{-4 x}{4 + x^2} dx$$, substitute $$u = 4 + x^2$$, $$du = 2 x dx$$, so $$x dx = \frac{du}{2}$$. 64. Rewrite integral: $$\int_0^2 \frac{-4 x}{4 + x^2} dx = -4 \int_0^2 \frac{x}{4 + x^2} dx = -4 \int_{u=4}^{8} \frac{1}{u} \cdot \frac{du}{2} = -2 \int_4^8 \frac{1}{u} du = -2 (\ln 8 - \ln 4) = -2 \ln 2$$ 65. For $$\int_0^2 \frac{6}{4 + x^2} dx$$: $$6 \int_0^2 \frac{1}{4 + x^2} dx = 6 \left[ \frac{1}{2} \arctan \frac{x}{2} \right]_0^2 = 3 (\arctan 1 - \arctan 0) = 3 \cdot \frac{\pi}{4} = \frac{3 \pi}{4}$$ 66. Sum all parts: $$-3 \ln 3 - 2 \ln 2 + \frac{3 \pi}{4} = \frac{3 \pi}{4} - \ln (3^3 \cdot 2^2) = \frac{3 \pi}{4} - \ln 108$$ --- **Final answers:** 1. $$\frac{53}{3} \ln 3 - \frac{26}{9}$$ 2(a). Stationary point at $$\left( \ln \frac{4}{3}, -16 \right)$$ 2(b). $$\int_{\ln 3}^{\ln 5} f(x) dx = \ln \frac{81}{4}$$ 3(a). $$a = 5.04$$ 3(b). Area $$= 2 \sqrt{2}$$ 4(a). $$f(x) = \frac{-3}{1 + x} + \frac{-4 x + 6}{4 + x^2}$$ 4(b). $$\int_0^2 f(x) dx = \frac{3 \pi}{4} - \ln 108$$