Evaluate the integral. \[ \int_{0}^{3} \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x \]

To evaluate the integral \[ I = \int_{0}^{3} \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} \, dx, \] we can begin by simplifying the integrand through partial fraction decomposition. We will express \[ \frac{x^2 + x + 1}{(x+1)^2 (x+2)} \] as a sum of simpler fractions. We want to find constants \( A \), \( B \), and \( C \) such that \[ \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+2}. \] Multiplying both sides by the denominator \( (x+1)^2 (x+2) \) gives: \[ x^2 + x + 1 = A(x+1)(x+2) + B(x+2) + C(x+1)^2. \] Expanding the right-hand side: 1. \( A(x+1)(x+2) = A(x^2 + 3x + 2) = Ax^2 + 3Ax + 2A \) 2. \( B(x+2) = Bx + 2B \) 3. \( C(x+1)^2 = C(x^2 + 2x + 1) = Cx^2 + 2Cx + C \) Combining these, we have: \[ (A+C)x^2 + (3A + B + 2C)x + (2A + 2B + C). \] Setting the coefficients equal to those in \( x^2 + x + 1 \): 1. For \( x^2 \): \( A + C = 1 \) 2. For \( x \): \( 3A + B + 2C = 1 \) 3. For the constant: \( 2A + 2B + C = 1 \) Now we can solve this system of equations. From the first equation, we can express \( C = 1 - A \). Substituting \( C \) into the other two equations gives: For the second equation: \[ 3A + B + 2(1 - A) = 1 \implies 3A + B + 2 - 2A = 1 \implies A + B = -1 \quad (1) \] For the third equation: \[ 2A + 2B + (1 - A) = 1 \implies 2A + 2B + 1 - A = 1 \implies A + 2B = 0 \quad (2) \] We now have a system of two equations: 1. \( A + B = -1 \) 2. \( A + 2B = 0 \) Subtract the first from the second: \[ (A + 2B) - (A + B) = 0 - (-1) \implies B = 1. \] Substituting \( B = 1 \) back into \( A + B = -1 \): \[ A + 1 = -1 \implies A = -2. \] Finally, substituting \( A \) back to find \( C \): \[ C = 1 - A = 1 - (-2) = 3. \] The partial fraction decomposition is: \[ \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{-2}{x+1} + \frac{1}{(x+1)^2} + \frac{3}{x+2}. \] Now, we can split the integral: \[ I = \int_{0}^{3} \left( \frac{-2}{x+1} + \frac{1}{(x+1)^2} + \frac{3}{x+2} \right) dx. \] Calculating each integral separately: 1. \[ \int \frac{-2}{x+1} \, dx = -2 \ln|x+1| + C. \] 2. \[ \int \frac{1}{(x+1)^2} \, dx = -\frac{1}{x+1} + C. \] 3. \[ \int \frac{3}{x+2} \, dx = 3 \ln|x+2| + C. \] Combining these, we have: \[ I = \