1. Problem: Simplify the rational expression $$\frac{12x^2 - 7x - 10}{3x + 2}$$. 2. Use polynomial division or factorization to simplify. 3. Factor numerator: $$12x^2 - 7x - 10 = (4x + 5)(3x - 2)$$. 4. Rewrite expression: $$\frac{(4x + 5)(3x - 2)}{3x + 2}$$. 5. Since denominator is $$3x + 2$$, no common factor with numerator, so expression cannot be simplified further. 6. Problem: Simplify $$\frac{x^6 - 1}{x - 1}$$. 7. Use formula for difference of powers: $$a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$$. 8. Here, $$a = x$$, $$b = 1$$, $$n = 6$$. 9. So, $$x^6 - 1 = (x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1)$$. 10. Cancel $$x - 1$$ in numerator and denominator. 11. Final answer: $$x^5 + x^4 + x^3 + x^2 + x + 1$$. 12. Problem: Simplify $$\frac{x^4 + x^2 + 1}{x^2 - x + 1}$$. 13. Try polynomial division or factorization. 14. Multiply denominator by $$x^2 + x + 1$$: $$(x^2 - x + 1)(x^2 + x + 1) = x^4 + x^2 + 1$$. 15. So numerator equals denominator times $$x^2 + x + 1$$. 16. Therefore, $$\frac{x^4 + x^2 + 1}{x^2 - x + 1} = x^2 + x + 1$$. 17. Problem: Evaluate the integral $$\int \frac{e^{2x} - 1 - e^{1 - 2x}}{e^x + 2} \, dx$$. 18. Let $$t = e^x$$, so $$dt = e^x dx = t dx$$, hence $$dx = \frac{dt}{t}$$. 19. Rewrite numerator: $$e^{2x} = (e^x)^2 = t^2$$, $$e^{1 - 2x} = e^1 \cdot e^{-2x} = e \cdot t^{-2}$$. 20. Integral becomes $$\int \frac{t^2 - 1 - e t^{-2}}{t + 2} \cdot \frac{dt}{t} = \int \frac{t^2 - 1 - e t^{-2}}{t(t + 2)} dt$$. 21. Multiply numerator and denominator inside integral: $$\frac{t^2 - 1 - e t^{-2}}{t(t + 2)} = \frac{t^2 - 1 - e t^{-2}}{t^2 + 2t}$$. 22. Multiply numerator terms by $$t^2$$ to clear denominator inside numerator: numerator becomes $$t^4 - t^2 - e$$. 23. So integral is $$\int \frac{t^4 - t^2 - e}{t^2 (t + 2)} dt$$. 24. Perform polynomial division or partial fractions to simplify. 25. Divide numerator $$t^4 - t^2 - e$$ by denominator $$t^3 + 2 t^2$$. 26. Quotient: $$t - 2$$, remainder: $$3 t^2 - e$$. 27. Rewrite integral as $$\int (t - 2) dt + \int \frac{3 t^2 - e}{t^2 (t + 2)} dt$$. 28. First integral: $$\frac{t^2}{2} - 2 t$$. 29. For second integral, use partial fractions: $$\frac{3 t^2 - e}{t^2 (t + 2)} = \frac{A}{t} + \frac{B}{t^2} + \frac{C}{t + 2}$$. 30. Solve for A, B, C: $$3 t^2 - e = A t (t + 2) + B (t + 2) + C t^2$$. 31. Equate coefficients: - For $$t^2$$: $$3 = A + C$$ - For $$t$$: $$0 = 2 A + B$$ - For constant: $$-e = 2 B$$ 32. From constant: $$B = -\frac{e}{2}$$. 33. From $$t$$: $$0 = 2 A - \frac{e}{2} \Rightarrow A = \frac{e}{4}$$. 34. From $$t^2$$: $$3 = \frac{e}{4} + C \Rightarrow C = 3 - \frac{e}{4}$$. 35. Integral becomes: $$\int \left( \frac{e}{4 t} - \frac{e}{2 t^2} + \frac{3 - \frac{e}{4}}{t + 2} \right) dt$$. 36. Integrate term by term: - $$\int \frac{e}{4 t} dt = \frac{e}{4} \ln |t|$$ - $$\int -\frac{e}{2 t^2} dt = \frac{e}{2 t}$$ - $$\int \frac{3 - \frac{e}{4}}{t + 2} dt = \left(3 - \frac{e}{4}\right) \ln |t + 2|$$ 37. Combine all parts: $$\frac{t^2}{2} - 2 t + \frac{e}{4} \ln |t| + \frac{e}{2 t} + \left(3 - \frac{e}{4}\right) \ln |t + 2| + C$$. 38. Substitute back $$t = e^x$$: $$\frac{e^{2x}}{2} - 2 e^x + \frac{e}{4} x + \frac{e}{2 e^x} + \left(3 - \frac{e}{4}\right) \ln (e^x + 2) + C$$. 39. Problem: Evaluate $$\int \frac{27^{1 + x} + 9^{1 - x}}{3^x} dx$$. 40. Express bases as powers of 3: - $$27 = 3^3$$, so $$27^{1 + x} = 3^{3(1 + x)} = 3^{3 + 3x}$$. - $$9 = 3^2$$, so $$9^{1 - x} = 3^{2(1 - x)} = 3^{2 - 2x}$$. 41. Rewrite integral: $$\int \frac{3^{3 + 3x} + 3^{2 - 2x}}{3^x} dx = \int (3^{3 + 3x - x} + 3^{2 - 2x - x}) dx = \int (3^{3 + 2x} + 3^{2 - 3x}) dx$$. 42. Simplify exponents: $$3^{3 + 2x} = 3^3 \cdot 3^{2x} = 27 \cdot (3^{2x})$$ $$3^{2 - 3x} = 3^2 \cdot 3^{-3x} = 9 \cdot (3^{-3x})$$ 43. Integral becomes: $$\int 27 \cdot 3^{2x} + 9 \cdot 3^{-3x} dx = 27 \int 3^{2x} dx + 9 \int 3^{-3x} dx$$. 44. Use formula $$\int a^{kx} dx = \frac{a^{kx}}{k \ln a} + C$$. 45. Compute each integral: - $$27 \int 3^{2x} dx = 27 \cdot \frac{3^{2x}}{2 \ln 3} = \frac{27}{2 \ln 3} 3^{2x}$$ - $$9 \int 3^{-3x} dx = 9 \cdot \frac{3^{-3x}}{-3 \ln 3} = -\frac{3}{\ln 3} 3^{-3x}$$ 46. Final answer: $$\frac{27}{2 \ln 3} 3^{2x} - \frac{3}{\ln 3} 3^{-3x} + C$$.