1. **Problem:** Evaluate $$\int \frac{4x + 2}{x^3 + x^2 - 2x} \, dx$$ using partial fractions. Step 1: Factor the denominator: $$x^3 + x^2 - 2x = x(x^2 + x - 2) = x(x+2)(x-1)$$ Step 2: Set up partial fractions: $$\frac{4x + 2}{x(x+2)(x-1)} = \frac{A}{x} + \frac{B}{x+2} + \frac{C}{x-1}$$ Step 3: Multiply both sides by the denominator: $$4x + 2 = A(x+2)(x-1) + Bx(x-1) + Cx(x+2)$$ Step 4: Expand and simplify: $$4x + 2 = A(x^2 + x - 2) + B(x^2 - x) + C(x^2 + 2x)$$ $$= (A + B + C)x^2 + (A - B + 2C)x - 2A$$ Step 5: Equate coefficients: - Coefficient of $$x^2$$: $$0 = A + B + C$$ - Coefficient of $$x$$: $$4 = A - B + 2C$$ - Constant term: $$2 = -2A$$ Step 6: Solve for $$A$$ from constant term: $$2 = -2A \Rightarrow A = -1$$ Step 7: Substitute $$A = -1$$ into the other equations: $$0 = -1 + B + C \Rightarrow B + C = 1$$ $$4 = -1 - B + 2C \Rightarrow -B + 2C = 5$$ Step 8: Solve system: From $$B + C = 1$$, $$B = 1 - C$$. Substitute into $$-B + 2C = 5$$: $$-(1 - C) + 2C = 5 \Rightarrow -1 + C + 2C = 5 \Rightarrow 3C = 6 \Rightarrow C = 2$$ Then $$B = 1 - 2 = -1$$. Step 9: Write integral with partial fractions: $$\int \left(\frac{-1}{x} + \frac{-1}{x+2} + \frac{2}{x-1}\right) dx = \int \left(-\frac{1}{x} - \frac{1}{x+2} + \frac{2}{x-1}\right) dx$$ Step 10: Integrate: $$= - \ln|x| - \ln|x+2| + 2 \ln|x-1| + C$$ --- 2. **Problem:** Evaluate $$\int \frac{x^4 + 3x^3 - 4x + 1}{x^3 + 4x^2 + 4x} \, dx$$ using partial fractions. Step 1: Factor denominator: $$x^3 + 4x^2 + 4x = x(x+2)^2$$ Step 2: Since numerator degree (4) ≥ denominator degree (3), perform polynomial division: Divide $$x^4 + 3x^3 - 4x + 1$$ by $$x^3 + 4x^2 + 4x$$. Leading term division: $$x^4 / x^3 = x$$. Multiply divisor by $$x$$: $$x(x^3 + 4x^2 + 4x) = x^4 + 4x^3 + 4x^2$$. Subtract: $$ (x^4 + 3x^3 - 4x + 1) - (x^4 + 4x^3 + 4x^2) = 0x^4 - x^3 - 4x^2 - 4x + 1$$ Step 3: Divide $$-x^3$$ by $$x^3$$ is $$-1$$. Multiply divisor by $$-1$$: $$-(x^3 + 4x^2 + 4x) = -x^3 - 4x^2 - 4x$$. Subtract: $$( -x^3 - 4x^2 - 4x + 1 ) - ( -x^3 - 4x^2 - 4x ) = 1$$ Remainder: $$1$$. Step 4: Write polynomial division result: $$\frac{x^4 + 3x^3 - 4x + 1}{x^3 + 4x^2 + 4x} = x - 1 + \frac{1}{x(x+2)^2}$$ Step 5: Decompose $$\frac{1}{x(x+2)^2}$$: $$\frac{1}{x(x+2)^2} = \frac{A}{x} + \frac{B}{x+2} + \frac{C}{(x+2)^2}$$ Step 6: Multiply both sides by $$x(x+2)^2$$: $$1 = A(x+2)^2 + B x (x+2) + C x$$ Step 7: Expand: $$1 = A(x^2 + 4x + 4) + B(x^2 + 2x) + C x = (A + B) x^2 + (4A + 2B + C) x + 4A$$ Step 8: Equate coefficients: - Coefficient of $$x^2$$: $$0 = A + B$$ - Coefficient of $$x$$: $$0 = 4A + 2B + C$$ - Constant term: $$1 = 4A$$ Step 9: Solve: $$4A = 1 \Rightarrow A = \frac{1}{4}$$ $$B = -A = -\frac{1}{4}$$ $$0 = 4 \times \frac{1}{4} + 2(-\frac{1}{4}) + C = 1 - \frac{1}{2} + C = \frac{1}{2} + C \Rightarrow C = -\frac{1}{2}$$ Step 10: Write integral: $$\int \left(x - 1 + \frac{\frac{1}{4}}{x} - \frac{\frac{1}{4}}{x+2} - \frac{\frac{1}{2}}{(x+2)^2}\right) dx$$ Step 11: Integrate terms: $$\int x \, dx = \frac{x^2}{2}, \quad \int (-1) dx = -x$$ $$\int \frac{1/4}{x} dx = \frac{1}{4} \ln |x|$$ $$\int -\frac{1/4}{x+2} dx = -\frac{1}{4} \ln |x+2|$$ $$\int -\frac{1/2}{(x+2)^2} dx = -\frac{1}{2} \int (x+2)^{-2} dx = -\frac{1}{2} \times \left(-\frac{1}{x+2}\right) = \frac{1}{2(x+2)}$$ Step 12: Final answer: $$\frac{x^2}{2} - x + \frac{1}{4} \ln |x| - \frac{1}{4} \ln |x+2| + \frac{1}{2(x+2)} + C$$ --- 3. **Problem:** Evaluate $$\int \frac{x^3}{(x^2 + 4)^2} \, dx$$ Step 1: Use substitution Let $$u = x^2 + 4 \Rightarrow du = 2x \, dx \Rightarrow x \, dx = \frac{du}{2}$$ Rewrite numerator: $$x^3 \, dx = x^2 (x \, dx) = (u - 4) \times \frac{du}{2}$$ Step 2: Change integral: $$\int \frac{x^3}{(x^2 + 4)^2} dx = \int \frac{(u - 4) \cdot \frac{du}{2}}{u^2} = \frac{1}{2} \int \frac{u - 4}{u^2} du = \frac{1}{2} \int \left(\frac{u}{u^2} - \frac{4}{u^2}\right) du = \frac{1}{2} \int \left(u^{-1} - 4 u^{-2}\right) du$$ Step 3: Integrate: $$\frac{1}{2} \left( \ln|u| + \frac{4}{u} \right) + C = \frac{1}{2} \ln|x^2 + 4| + \frac{2}{x^2 + 4} + C$$ --- **Final answers:** 1. $$- \ln|x| - \ln|x+2| + 2 \ln|x-1| + C$$ 2. $$\frac{x^2}{2} - x + \frac{1}{4} \ln|x| - \frac{1}{4} \ln|x+2| + \frac{1}{2(x+2)} + C$$ 3. $$\frac{1}{2} \ln|x^2 + 4| + \frac{2}{x^2 + 4} + C$$