1. Evaluate the integral $$\int \frac{3y}{5y^2 + 4} dy$$. Step 1: Substitute to simplify the integral. Let $$u = 5y^2 + 4$$, then $$du = 10y dy$$, so $$y dy = \frac{du}{10}$$. Step 2: Rewrite the integral in terms of $$u$$: $$\int \frac{3y}{u} dy = 3 \int \frac{y dy}{u} = 3 \int \frac{1}{u} \cdot y dy = 3 \int \frac{1}{u} \cdot \frac{du}{10} = \frac{3}{10} \int \frac{1}{u} du$$. Step 3: Integrate: $$\frac{3}{10} \int \frac{1}{u} du = \frac{3}{10} \ln|u| + C = \frac{3}{10} \ln|5y^2 + 4| + C$$. 2. Evaluate the integral $$\int \frac{x^2 - 3x + 4}{x^3 - 4x^2} dx$$. Step 1: Factor the denominator: $$x^3 - 4x^2 = x^2(x - 4)$$. Step 2: Use partial fractions decomposition: Set $$\frac{x^2 - 3x + 4}{x^2 (x - 4)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x -4}$$. Step 3: Multiply both sides by denominator for equation: $$x^2 - 3x + 4 = A x (x - 4) + B (x - 4) + C x^2$$. Step 4: Expand right side: $$A x^2 - 4 A x + B x - 4 B + C x^2 = (A + C) x^2 + (-4A + B) x - 4 B$$. Step 5: Equate coefficients: For $$x^2$$: $$1 = A + C$$ For $$x$$: $$-3 = -4 A + B$$ For constant: $$4 = -4 B$$ Step 6: Solve for B: $$4 = -4 B \implies B = -1$$. Step 7: Plug $$B=-1$$ into $$x$$ coefficient equation: $$-3 = -4 A -1 \implies -4 A = -2 \implies A = \frac{1}{2}$$. Step 8: Find C from $$x^2$$ coefficient: $$1 = A + C = \frac{1}{2} + C \implies C = \frac{1}{2}$$. Step 9: Rewrite integral using partial fractions: $$\int \left( \frac{1/2}{x} - \frac{1}{x^2} + \frac{1/2}{x -4} \right) dx = \frac{1}{2} \int \frac{dx}{x} - \int \frac{dx}{x^2} + \frac{1}{2} \int \frac{dx}{x -4}$$. Step 10: Integrate each term: $$\frac{1}{2} \ln |x| + \frac{1}{x} + \frac{1}{2} \ln |x -4| + C$$. 3. Evaluate the definite integral $$\int_{1}^{2} \frac{(\ln x)^2}{x^3} dx$$. Step 1: Use substitution. Let $$t = \ln x \implies dt = \frac{1}{x} dx \implies dx = x dt$$. Step 2: Rewrite the integral: $$\frac{(\ln x)^2}{x^3} dx = \frac{t^2}{x^3} dx = t^2 \frac{1}{x^3} x dt = t^2 \frac{dt}{x^2} = t^2 e^{-2t} dt \quad \text{(since } x = e^t\text{)}$$. Step 3: Change limits: When $$x=1$$, $$t=\ln 1 = 0$$. When $$x=2$$, $$t=\ln 2$$. Step 4: The integral becomes: $$\int_0^{\ln 2} t^2 e^{-2t} dt$$. Step 5: Integrate by parts twice to evaluate: Let $$I = \int t^2 e^{-2t} dt$$. First integration by parts: - $$u = t^2, \quad dv = e^{-2t} dt$$ - $$du = 2t dt, \quad v = -\frac{1}{2} e^{-2t}$$ Then, $$I = -\frac{1}{2} t^2 e^{-2t} + \int t (e^{-2t}) dt$$ Step 6: Integrate $$\int t e^{-2t} dt$$ by parts again: - $$u = t, dv = e^{-2t} dt$$ - $$du = dt, v = -\frac{1}{2} e^{-2t}$$ So, $$\int t e^{-2t} dt = -\frac{1}{2} t e^{-2t} + \frac{1}{2} \int e^{-2t} dt = -\frac{1}{2} t e^{-2t} - \frac{1}{4} e^{-2t} + C$$ Step 7: Substitute back: $$I = -\frac{1}{2} t^2 e^{-2t} - \frac{1}{2} t e^{-2t} - \frac{1}{4} e^{-2t} + C = -\frac{e^{-2t}}{4} (2 t^2 + 2 t + 1) + C$$. Step 8: Evaluate definite integral from 0 to $$\ln 2$$: $$\left[-\frac{e^{-2t}}{4} (2 t^2 + 2 t + 1) \right]_0^{\ln 2} = -\frac{e^{-2 \ln 2}}{4} (2 (\ln 2)^2 + 2 \ln 2 + 1) + \frac{e^{0}}{4} (0 + 0 + 1)$$. Step 9: Simplify terms: Recall $$e^{-2 \ln 2} = e^{\ln 2^{-2}} = 2^{-2} = \frac{1}{4}$$. So, $$= -\frac{1}{4} \cdot \frac{1}{4} (2 (\ln 2)^2 + 2 \ln 2 + 1) + \frac{1}{4} = -\frac{1}{16} (2 (\ln 2)^2 + 2 \ln 2 + 1) + \frac{1}{4}$$. Step 10: Final result: $$\frac{1}{4} - \frac{1}{16} (2 (\ln 2)^2 + 2 \ln 2 + 1)$$. --- Final answers: 1. $$\int \frac{3y}{5y^2 + 4} dy = \frac{3}{10} \ln |5y^2 + 4| + C$$ 2. $$\int \frac{x^2 - 3x + 4}{x^3 - 4x^2} dx = \frac{1}{2} \ln|x| + \frac{1}{x} + \frac{1}{2} \ln|x-4| + C$$ 3. $$\int_1^2 \frac{(\ln x)^2}{x^3} dx = \frac{1}{4} - \frac{1}{16} \bigl(2 (\ln 2)^2 + 2 \ln 2 + 1\bigr)$$