2. Evaluate \( \int \frac{5 x^{2}-3 x+2}{x^{3}-2 x^{2}} d x \)

To evaluate the integral \( \int \frac{5 x^{2}-3 x+2}{x^{3}-2 x^{2}} d x \), we start by simplifying the integrand: First, factor the denominator: \[ x^3 - 2x^2 = x^2(x - 2). \] Thus, the integrand becomes: \[ \frac{5x^2 - 3x + 2}{x^2(x - 2)}. \] Next, we can perform polynomial long division or partial fraction decomposition. The degree of the numerator is less than that of the denominator after factoring out \( x^2 \). So, we can write: \[ \frac{5x^2 - 3x + 2}{x^2(x - 2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 2}, \] where \( A \), \( B \), and \( C \) are constants to be determined. Multiplying through by \( x^2(x - 2) \) gives: \[ 5x^2 - 3x + 2 = A(x)(x - 2) + B(x - 2) + C(x^2). \] Expanding the right side: \[ A(x^2 - 2x) + B(x - 2) + Cx^2 = (A + C)x^2 + (-2A + B)x - 2B. \] Setting this equal to \( 5x^2 - 3x + 2 \), we equate coefficients: 1. \( A + C = 5 \) 2. \( -2A + B = -3 \) 3. \( -2B = 2 \) (which implies \( B = -1 \)) Now, substituting \( B = -1 \) into the second equation: \[ -2A - 1 = -3 \implies -2A = -2 \implies A = 1. \] Now substituting \( A = 1 \) into the first equation: \[ 1 + C = 5 \implies C = 4. \] So we have \( A = 1 \), \( B = -1 \), and \( C = 4 \). Thus, the partial fraction decomposition is: \[ \frac{1}{x} - \frac{1}{x^2} + \frac{4}{x - 2}. \] Now, we can integrate term by term: \[ \int \left( \frac{1}{x} - \frac{1}{x^2} + \frac{4}{x - 2} \right) dx = \ln |x| + \frac{1}{x} + 4 \ln |x - 2| + C, \] where \( C \) is the constant of integration. Putting it all together, the evaluated integral is: \[ \int \frac{5 x^{2}-3 x+2}{x^{3}-2 x^{2}} d x = \ln |x| + \frac{1}{x} + 4 \ln |x - 2| + C. \]