1. **Problem 1:** Evaluate $$\int \frac{x^3}{(x+1)^2} \, dx$$ by expressing the integrand in partial fractions. 2. **Problem 2:** Evaluate $$\int \frac{11x - 66}{x^2 - x - 2} \, dx$$ by expressing the integrand in partial fractions. 3. **Problem 3:** Evaluate $$\int \frac{1}{(x-5)(x+1)} \, dx$$ by expressing the integrand in partial fractions. 4. **Problem 4:** Evaluate $$\int \frac{8x}{(x-1)^2 (x+1)} \, dx$$ by expressing the integrand in partial fractions. --- ### Step-by-step solutions: ### 1. $$\int \frac{x^3}{(x+1)^2} \, dx$$ 1. Express the integrand as $$\frac{x^3}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + Cx^2 + Dx + E$$ because the numerator degree is higher than denominator. 2. Multiply both sides by $$(x+1)^2$$: $$x^3 = A(x+1) + B + (Cx^2 + Dx + E)(x+1)^2$$ 3. Expand and equate coefficients to solve for constants. 4. Alternatively, perform polynomial division first: $$\frac{x^3}{(x+1)^2} = Q(x) + \frac{R(x)}{(x+1)^2}$$ Since $$(x+1)^2 = x^2 + 2x + 1$$, divide $x^3$ by $x^2 + 2x + 1$: - Leading term: $x^3 / x^2 = x$ - Multiply: $x(x^2 + 2x + 1) = x^3 + 2x^2 + x$ - Subtract: $(x^3 - x^3) + (0 - 2x^2) + (0 - x) = -2x^2 - x$ - Next term: $-2x^2 / x^2 = -2$ - Multiply: $-2(x^2 + 2x + 1) = -2x^2 - 4x - 2$ - Subtract: $(-2x^2 - x) - (-2x^2 - 4x - 2) = 0 + 3x + 2$ So, $$\frac{x^3}{(x+1)^2} = x - 2 + \frac{3x + 2}{(x+1)^2}$$ 5. Now express $$\frac{3x + 2}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2}$$ Multiply both sides by $$(x+1)^2$$: $$3x + 2 = A(x+1) + B$$ Set $x = -1$: $$3(-1) + 2 = A(0) + B \Rightarrow -3 + 2 = B \Rightarrow B = -1$$ For $x=0$: $$3(0) + 2 = A(1) + B \Rightarrow 2 = A -1 \Rightarrow A = 3$$ 6. So, $$\frac{3x + 2}{(x+1)^2} = \frac{3}{x+1} - \frac{1}{(x+1)^2}$$ 7. The integral becomes: $$\int \left(x - 2 + \frac{3}{x+1} - \frac{1}{(x+1)^2}\right) dx$$ 8. Integrate term by term: $$\int x \, dx = \frac{x^2}{2}$$ $$\int -2 \, dx = -2x$$ $$\int \frac{3}{x+1} \, dx = 3 \ln|x+1|$$ $$\int -\frac{1}{(x+1)^2} \, dx = \frac{1}{x+1}$$ 9. Final answer: $$\frac{x^2}{2} - 2x + 3 \ln|x+1| + \frac{1}{x+1} + C$$ --- ### 2. $$\int \frac{11x - 66}{x^2 - x - 2} \, dx$$ 1. Factor denominator: $$x^2 - x - 2 = (x - 2)(x + 1)$$ 2. Express integrand as: $$\frac{11x - 66}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1}$$ 3. Multiply both sides by denominator: $$11x - 66 = A(x + 1) + B(x - 2)$$ 4. Expand: $$11x - 66 = A x + A + B x - 2B = (A + B) x + (A - 2B)$$ 5. Equate coefficients: - Coefficient of $x$: $11 = A + B$ - Constant term: $-66 = A - 2B$ 6. Solve system: From first: $A = 11 - B$ Substitute into second: $$-66 = (11 - B) - 2B = 11 - 3B \Rightarrow -66 - 11 = -3B \Rightarrow -77 = -3B \Rightarrow B = \frac{77}{3}$$ Then: $$A = 11 - \frac{77}{3} = \frac{33}{3} - \frac{77}{3} = -\frac{44}{3}$$ 7. Integral becomes: $$\int \left( \frac{-44/3}{x - 2} + \frac{77/3}{x + 1} \right) dx = -\frac{44}{3} \int \frac{1}{x - 2} dx + \frac{77}{3} \int \frac{1}{x + 1} dx$$ 8. Integrate: $$-\frac{44}{3} \ln|x - 2| + \frac{77}{3} \ln|x + 1| + C$$ --- ### 3. $$\int \frac{1}{(x-5)(x+1)} \, dx$$ 1. Express integrand as: $$\frac{1}{(x-5)(x+1)} = \frac{A}{x-5} + \frac{B}{x+1}$$ 2. Multiply both sides by denominator: $$1 = A(x+1) + B(x-5) = (A + B) x + (A - 5B)$$ 3. Equate coefficients: - Coefficient of $x$: $0 = A + B$ - Constant term: $1 = A - 5B$ 4. From first: $A = -B$ Substitute into second: $$1 = -B - 5B = -6B \Rightarrow B = -\frac{1}{6}$$ Then: $$A = -B = \frac{1}{6}$$ 5. Integral becomes: $$\int \left( \frac{1/6}{x-5} - \frac{1/6}{x+1} \right) dx = \frac{1}{6} \int \frac{1}{x-5} dx - \frac{1}{6} \int \frac{1}{x+1} dx$$ 6. Integrate: $$\frac{1}{6} \ln|x-5| - \frac{1}{6} \ln|x+1| + C = \frac{1}{6} \ln \left| \frac{x-5}{x+1} \right| + C$$ --- ### 4. $$\int \frac{8x}{(x-1)^2 (x+1)} \, dx$$ 1. Express integrand as: $$\frac{8x}{(x-1)^2 (x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1}$$ 2. Multiply both sides by denominator: $$8x = A(x-1)(x+1) + B(x+1) + C(x-1)^2$$ 3. Expand: - $A(x-1)(x+1) = A(x^2 - 1)$ - $B(x+1) = Bx + B$ - $C(x-1)^2 = C(x^2 - 2x + 1)$ Sum: $$8x = A x^2 - A + B x + B + C x^2 - 2 C x + C$$ Group terms: $$8x = (A + C) x^2 + (B - 2C) x + (-A + B + C)$$ 4. Equate coefficients: - $x^2$: $0 = A + C$ - $x$: $8 = B - 2C$ - Constant: $0 = -A + B + C$ 5. From first: $C = -A$ 6. Substitute into second: $$8 = B - 2(-A) = B + 2A$$ 7. Substitute into third: $$0 = -A + B + (-A) = B - 2A$$ 8. From third: $B = 2A$ 9. Substitute $B=2A$ into second: $$8 = 2A + 2A = 4A \Rightarrow A = 2$$ 10. Then $B = 2A = 4$, and $C = -A = -2$ 11. Integral becomes: $$\int \left( \frac{2}{x-1} + \frac{4}{(x-1)^2} - \frac{2}{x+1} \right) dx$$ 12. Integrate term by term: $$\int \frac{2}{x-1} dx = 2 \ln|x-1|$$ $$\int \frac{4}{(x-1)^2} dx = 4 \int (x-1)^{-2} dx = 4 \left(-\frac{1}{x-1}\right) = -\frac{4}{x-1}$$ $$\int -\frac{2}{x+1} dx = -2 \ln|x+1|$$ 13. Final answer: $$2 \ln|x-1| - \frac{4}{x-1} - 2 \ln|x+1| + C$$