1. Problem: Calculate the integrals given in parts a, b, c, and d. 2. Part a: \(\int \sqrt{16x} \sin(1 + x^{3/2}) \, dx\) - Rewrite \(\sqrt{16x} = 4\sqrt{x} = 4x^{1/2}\). - Let \(u = 1 + x^{3/2}\), then \(du = \frac{3}{2} x^{1/2} dx\). - Solve for \(x^{1/2} dx = \frac{2}{3} du\). - Substitute: \(4x^{1/2} \sin(u) dx = 4 \sin(u) \cdot x^{1/2} dx = 4 \sin(u) \cdot \frac{2}{3} du = \frac{8}{3} \sin(u) du\). - Integral becomes \(\int \frac{8}{3} \sin(u) du = \frac{8}{3} \int \sin(u) du = -\frac{8}{3} \cos(u) + C\). - Substitute back \(u = 1 + x^{3/2}\). - Final answer: \(-\frac{8}{3} \cos(1 + x^{3/2}) + C\). 3. Part b: \(\int_5^{12} x^3 \ln(6x) \, dx\) - Use integration by parts: let \(u = \ln(6x)\), \(dv = x^3 dx\). - Then \(du = \frac{1}{x} dx\), \(v = \frac{x^4}{4}\). - Integration by parts formula: \(\int u dv = uv - \int v du\). - Compute: \(= \left. \frac{x^4}{4} \ln(6x) \right|_5^{12} - \int_5^{12} \frac{x^4}{4} \cdot \frac{1}{x} dx = \left. \frac{x^4}{4} \ln(6x) \right|_5^{12} - \frac{1}{4} \int_5^{12} x^3 dx\). - Evaluate \(\int_5^{12} x^3 dx = \left. \frac{x^4}{4} \right|_5^{12} = \frac{12^4 - 5^4}{4} = \frac{20736 - 625}{4} = \frac{20111}{4}\). - Substitute back: \(= \frac{12^4}{4} \ln(72) - \frac{5^4}{4} \ln(30) - \frac{1}{4} \cdot \frac{20111}{4} = 10368 \ln(72) - 156.25 \ln(30) - \frac{20111}{16}\). 4. Part c: \(\int \left( \frac{1 - x^{-2}}{x^{1/2} - x^{-1/2}} - \frac{2}{x^{3/2}} + \frac{x^{-2} - x}{x^{1/2} - x^{-1/2}} \right) dx\) - Combine fractions over common denominator \(x^{1/2} - x^{-1/2}\). - Simplify numerator: \(\frac{1 - x^{-2} + x^{-2} - x}{x^{1/2} - x^{-1/2}} - \frac{2}{x^{3/2}} = \frac{1 - x}{x^{1/2} - x^{-1/2}} - \frac{2}{x^{3/2}}\). - Note \(x^{1/2} - x^{-1/2} = \frac{x - 1}{x^{1/2}}\). - So \(\frac{1 - x}{x^{1/2} - x^{-1/2}} = \frac{1 - x}{\frac{x - 1}{x^{1/2}}} = (1 - x) \cdot \frac{x^{1/2}}{x - 1} = -x^{1/2}\). - The integral becomes \(\int (-x^{1/2} - 2x^{-3/2}) dx = \int (-x^{1/2}) dx - 2 \int x^{-3/2} dx\). - Integrate: \(\int x^{1/2} dx = \frac{2}{3} x^{3/2}\), \(\int x^{-3/2} dx = -2 x^{-1/2}\). - So integral is \(- \frac{2}{3} x^{3/2} - 2 (-2 x^{-1/2}) + C = - \frac{2}{3} x^{3/2} + 4 x^{-1/2} + C\). 5. Part d: \(\int \frac{x + 7}{2x^2 + 9x - 18} dx\) - Factor denominator: \(2x^2 + 9x - 18 = (2x - 3)(x + 6)\). - Use partial fractions: \(\frac{x + 7}{(2x - 3)(x + 6)} = \frac{A}{2x - 3} + \frac{B}{x + 6}\). - Multiply both sides by denominator: \(x + 7 = A(x + 6) + B(2x - 3)\). - Expand: \(x + 7 = A x + 6A + 2 B x - 3 B = (A + 2B) x + (6A - 3B)\). - Equate coefficients: \(A + 2B = 1\), \(6A - 3B = 7\). - Solve system: From first: \(A = 1 - 2B\). Substitute into second: \(6(1 - 2B) - 3B = 7 \Rightarrow 6 - 12B - 3B = 7 \Rightarrow -15B = 1 \Rightarrow B = -\frac{1}{15}\). Then \(A = 1 - 2(-\frac{1}{15}) = 1 + \frac{2}{15} = \frac{17}{15}\). - Integral becomes: \(\int \frac{17/15}{2x - 3} dx + \int \frac{-1/15}{x + 6} dx = \frac{17}{15} \int \frac{1}{2x - 3} dx - \frac{1}{15} \int \frac{1}{x + 6} dx\). - Integrate: \(\int \frac{1}{2x - 3} dx = \frac{1}{2} \ln|2x - 3| + C\), \(\int \frac{1}{x + 6} dx = \ln|x + 6| + C\). - Final answer: \(\frac{17}{15} \cdot \frac{1}{2} \ln|2x - 3| - \frac{1}{15} \ln|x + 6| + C = \frac{17}{30} \ln|2x - 3| - \frac{1}{15} \ln|x + 6| + C\). Final answers: - a) \(-\frac{8}{3} \cos(1 + x^{3/2}) + C\) - b) \(10368 \ln(72) - 156.25 \ln(30) - \frac{20111}{16}\) - c) \(- \frac{2}{3} x^{3/2} + 4 x^{-1/2} + C\) - d) \(\frac{17}{30} \ln|2x - 3| - \frac{1}{15} \ln|x + 6| + C\)