The evaluation of the integral \( I = \int \frac{x^2 + 1}{(x - 1)^2 (x + 3)} \, dx \) requires partial fraction decomposition. The integrand is expressed as: \[ \frac{x^2 + 1}{(x - 1)^2 (x + 3)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{x + 3}. \] Multiplying by the common denominator yields: \[ x^2 + 1 = A(x - 1)(x + 3) + B(x + 3) + C(x - 1)^2. \] Expanding and collecting terms by powers of \(x\): \[ x^2 + 1 = (A + C)x^2 + (2A - 2C + B)x + (-3A + 3B + C). \] Equating coefficients of like powers of \(x\) provides a system of linear equations: 1. \(x^2\): \(A + C = 1\) 2. \(x\): \(2A - 2C + B = 0\) 3. Constant: \(-3A + 3B + C = 1\) Solving this system yields \(A = \frac{3}{8}\), \(B = \frac{1}{2}\), and \(C = \frac{5}{8}\). Thus, the decomposition is: \[ \frac{x^2 + 1}{(x - 1)^2 (x + 3)} = \frac{\frac{3}{8}}{x - 1} + \frac{\frac{1}{2}}{(x - 1)^2} + \frac{\frac{5}{8}}{x + 3}. \] The integral is then computed as: \[ I = \frac{3}{8} \int \frac{1}{x - 1} \, dx + \frac{1}{2} \int \frac{1}{(x - 1)^2} \, dx + \frac{5}{8} \int \frac{1}{x + 3} \, dx. \] The final result is: \[ I = \frac{3}{8} \ln |x - 1| - \frac{1}{2(x - 1)} + \frac{5}{8} \ln |x + 3| + C. \]