9.2. \( \int \frac{x^{2}-1}{x(x-1)^{2}} d x \)

To solve the integral \( \int \frac{x^{2}-1}{x(x-1)^{2}} d x \), we can start with partial fraction decomposition. We express: \[ \frac{x^{2}-1}{x(x-1)^{2}} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^{2}} \] Multiplying through by the common denominator \( x(x-1)^{2} \) gives: \[ x^{2}-1 = A(x-1)^{2} + Bx(x-1) + Cx \] Expanding the right side: \[ A(x^2 - 2x + 1) + B(x^2 - x) + Cx = Ax^2 - 2Ax + A + Bx^2 - Bx + Cx \] Combine like terms: \[ (A + B)x^2 + (-2A - B + C)x + A \] Matching coefficients from both sides, we have: 1. \( A + B = 1 \) 2. \( -2A - B + C = 0 \) 3. \( A = -1 \) From \( A = -1 \), substituting in the first equation gives \( -1 + B = 1 \) leading to \( B = 2 \). Then substituting \( A \) and \( B \) into the second equation: \[ -2(-1) - 2 + C = 0 \Rightarrow 2 - 2 + C = 0 \Rightarrow C = 0 \] Thus, we have \( A = -1 \), \( B = 2 \), and \( C = 0 \). Therefore, we rewrite the integral: \[ \frac{x^{2}-1}{x(x-1)^{2}} = \frac{-1}{x} + \frac{2}{x-1} \] Now we can integrate each term: \[ \int \left( -\frac{1}{x} + \frac{2}{x-1} \right) dx = -\ln |x| + 2\ln |x-1| + C \] So the final result is: \[ \int \frac{x^{2}-1}{x(x-1)^{2}} d x = -\ln |x| + 2\ln |x-1| + C \]