1. **State the problem:** We need to express the integrand $$\frac{x^3}{(x+1)^2}$$ in partial fractions and then evaluate the integral $$\int \frac{x^3}{(x+1)^2} \, dx$$. 2. **Set up the partial fraction decomposition:** Since the denominator is $(x+1)^2$, the partial fractions take the form: $$\frac{x^3}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + P(x)$$ where $P(x)$ is a polynomial because the degree of the numerator (3) is greater than the degree of the denominator (2). 3. **Perform polynomial division:** Divide $x^3$ by $(x+1)^2 = x^2 + 2x + 1$. - Divide $x^3$ by $x^2$ to get $x$. - Multiply $(x+1)^2$ by $x$: $x(x^2 + 2x + 1) = x^3 + 2x^2 + x$. - Subtract: $x^3 - (x^3 + 2x^2 + x) = -2x^2 - x$. - Divide $-2x^2$ by $x^2$ to get $-2$. - Multiply $(x+1)^2$ by $-2$: $-2(x^2 + 2x + 1) = -2x^2 - 4x - 2$. - Subtract: $(-2x^2 - x) - (-2x^2 - 4x - 2) = 3x + 2$. So the division gives: $$x^3 = (x - 2)(x+1)^2 + 3x + 2$$ 4. **Rewrite the integrand:** $$\frac{x^3}{(x+1)^2} = x - 2 + \frac{3x + 2}{(x+1)^2}$$ 5. **Decompose the remainder term:** $$\frac{3x + 2}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2}$$ Multiply both sides by $(x+1)^2$: $$3x + 2 = A(x+1) + B$$ 6. **Solve for A and B:** - Expand: $3x + 2 = A x + A + B$ - Equate coefficients: - Coefficient of $x$: $3 = A$ - Constant term: $2 = A + B = 3 + B \Rightarrow B = -1$ 7. **Rewrite the integrand fully:** $$\frac{x^3}{(x+1)^2} = x - 2 + \frac{3}{x+1} - \frac{1}{(x+1)^2}$$ 8. **Integrate term-by-term:** $$\int \frac{x^3}{(x+1)^2} \, dx = \int (x - 2) \, dx + \int \frac{3}{x+1} \, dx - \int \frac{1}{(x+1)^2} \, dx$$ - $$\int (x - 2) \, dx = \frac{x^2}{2} - 2x + C_1$$ - $$\int \frac{3}{x+1} \, dx = 3 \ln|x+1| + C_2$$ - $$\int \frac{1}{(x+1)^2} \, dx = \int (x+1)^{-2} \, dx = - (x+1)^{-1} + C_3 = -\frac{1}{x+1} + C_3$$ 9. **Combine all results:** $$\int \frac{x^3}{(x+1)^2} \, dx = \frac{x^2}{2} - 2x + 3 \ln|x+1| + \frac{1}{x+1} + C$$ where $C$ is the constant of integration. **Final answer:** $$\boxed{\int \frac{x^3}{(x+1)^2} \, dx = \frac{x^2}{2} - 2x + 3 \ln|x+1| + \frac{1}{x+1} + C}$$