1. Evaluate the integral $$\int \frac{3y}{y^2 + 4} \; dy$$ Step 1: Recognize the form suitable for substitution. Let $$u = y^2 + 4$$. Step 2: Then $$\frac{du}{dy} = 2y \implies du = 2y \; dy$$. Step 3: Rewrite the integral in terms of $$u$$: $$\int \frac{3y}{y^2 + 4} dy = \int \frac{3y}{u} dy$$. Step 4: Express $$y dy$$ in terms of $$du$$: $$y dy = \frac{du}{2}$$, so the integral becomes $$\int \frac{3}{u} \cdot \frac{du}{2} = \frac{3}{2} \int \frac{1}{u} du$$. Step 5: Integrate: $$\frac{3}{2} \ln|u| + C = \frac{3}{2} \ln|y^2 + 4| + C$$. 2. Evaluate the integral $$\int \frac{x^2 - 3x + 4}{x^4 - 4x^2} \; dx$$ Step 1: Factor the denominator: $$x^4 - 4x^2 = x^2(x^2 - 4) = x^2(x-2)(x+2)$$. Step 2: Partial fraction decomposition: let $$\frac{x^2 - 3x + 4}{x^2(x-2)(x+2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-2} + \frac{D}{x+2}$$. Step 3: Multiply both sides by the denominator: $$x^2 - 3x + 4 = A x (x-2)(x+2) + B (x-2)(x+2) + C x^2 (x+2) + D x^2 (x-2)$$. Step 4: Expand and collect terms, solve for constants: Evaluating at $$x=0$$: $$4 = B(-2)(2) = -4B \implies B = -1$$ At $$x=2$$: $$4 - 6 + 4 = 4 = C (4)(4) = 16 C \implies C = \frac{1}{4}$$ At $$x=-2$$: $$4 + 6 + 4 =14 = D (4)(-4) = -16 D \implies D = -\frac{14}{16} = -\frac{7}{8}$$ Use $$x=1$$ to solve for $$A$$: $$1 - 3 + 4 = 2 = A(1)(-1)(3) + B (-1)(3) + C (1)^2 (3) + D (1)^2 (-1)$$ $$2 = -3 A + (-1)(-1)(3) + \frac{1}{4} \cdot 3 + \left(-\frac{7}{8}\right) (-1)$$ $$2 = -3A + 3 + \frac{3}{4} + \frac{7}{8}$$ Sum constants: $$3 + 0.75 + 0.875 = 4.625$$ $$2 = -3 A + 4.625 \implies -3 A = 2 - 4.625 = -2.625 \implies A = \frac{2.625}{3} = 0.875 = \frac{7}{8}$$ Step 5: Integral becomes: $$\int \left( \frac{7}{8 x} - \frac{1}{x^2} + \frac{1}{4 (x-2)} - \frac{7}{8 (x+2)} \right) dx$$ Step 6: Integrate term-by-term: $$\frac{7}{8} \ln|x| + \frac{1}{x} + \frac{1}{4} \ln|x-2| - \frac{7}{8} \ln|x+2| + C$$ 3. Evaluate the definite integral $$\int_1^2 \frac{(\ln x)^2}{x^2} dx$$ Step 1: Make substitution: $$t = \ln x$$ so $$x = e^t$$, and $$dx = e^t dt$$ Step 2: Change limits: when $$x=1$$, $$t=0$$; when $$x=2$$, $$t=\ln 2$$ Step 3: Express integrand: $$\frac{(\ln x)^2}{x^2} dx = \frac{t^2}{e^{2 t}} e^t dt = t^2 e^{-t} dt$$ Step 4: Integral becomes: $$\int_0^{\ln 2} t^2 e^{-t} dt$$ Step 5: Use integration by parts twice: Let $$I = \int t^2 e^{-t} dt$$ First integration by parts: $$u = t^2 \implies du = 2t dt$$ $$dv = e^{-t} dt \implies v = -e^{-t}$$ $$I = -t^2 e^{-t} + 2 \int t e^{-t} dt$$ Second integration by parts on $$\int t e^{-t} dt$$: $$u = t \implies du = dt$$ $$dv = e^{-t} dt \implies v = -e^{-t}$$ $$\int t e^{-t} dt = -t e^{-t} + \int e^{-t} dt = -t e^{-t} - e^{-t} + C$$ Step 6: Substitute back: $$I = -t^2 e^{-t} + 2 (-t e^{-t} - e^{-t}) + C = -t^2 e^{-t} - 2 t e^{-t} - 2 e^{-t} + C$$ Step 7: Evaluate definite integral: $$I \bigg|_0^{\ln 2} = \left[-t^2 e^{-t} - 2 t e^{-t} - 2 e^{-t}\right]_0^{\ln 2}$$ Step 8: Calculate at $$t=\ln 2$$: $$e^{-\ln 2} = \frac{1}{2}, \quad t^2 = (\ln 2)^2, \quad t = \ln 2$$ $$- (\ln 2)^2 \cdot \frac{1}{2} - 2 \cdot \ln 2 \cdot \frac{1}{2} - 2 \cdot \frac{1}{2} = - \frac{(\ln 2)^2}{2} - \ln 2 -1$$ Step 9: Calculate at $$t=0$$: $$- 0 - 0 - 2 e^0 = -2$$ Step 10: Result: $$\left(- \frac{(\ln 2)^2}{2} - \ln 2 -1 \right) - (-2) = - \frac{(\ln 2)^2}{2} - \ln 2 -1 + 2 = - \frac{(\ln 2)^2}{2} - \ln 2 + 1$$ Final answers: 1. $$\frac{3}{2} \ln|y^2 + 4| + C$$ 2. $$\frac{7}{8} \ln|x| + \frac{1}{x} + \frac{1}{4} \ln|x-2| - \frac{7}{8} \ln|x+2| + C$$ 3. $$1 - \ln 2 - \frac{(\ln 2)^2}{2}$$