2. \( \int_{1}^{3} \frac{2 x-1}{x^{2}-2 x-3} d x= \)

To evaluate the integral \( \int_{1}^{3} \frac{2 x-1}{x^{2}-2 x-3} \, dx \), we first factor the denominator: \[ x^{2}-2x-3 = (x-3)(x+1). \] Now, we can express the function in terms of partial fractions: \[ \frac{2x-1}{(x-3)(x+1)} = \frac{A}{x-3} + \frac{B}{x+1}. \] To find \(A\) and \(B\), we set up the equation: \[ 2x - 1 = A(x + 1) + B(x - 3). \] Expanding and combining like terms, we have: \[ 2x - 1 = (A + B)x + (A - 3B). \] Setting the coefficients equal gives us the system of equations: 1. \( A + B = 2 \) 2. \( A - 3B = -1 \) Solving this system: From equation 1, we can express \( A = 2 - B \). Substituting into equation 2: \[ (2 - B) - 3B = -1 \implies 2 - 4B = -1 \implies -4B = -3 \implies B = \frac{3}{4}. \] Substituting back to find \( A \): \[ A + \frac{3}{4} = 2 \implies A = 2 - \frac{3}{4} = \frac{8}{4} - \frac{3}{4} = \frac{5}{4}. \] Thus, we have: \[ \frac{2x - 1}{(x-3)(x+1)} = \frac{5/4}{x-3} + \frac{3/4}{x+1}. \] Now, we can integrate term by term: \[ \int_{1}^{3} \frac{2x-1}{(x-3)(x+1)} \, dx = \int_{1}^{3} \left( \frac{5/4}{x-3} + \frac{3/4}{x+1} \right) \, dx. \] This gives: \[ = \frac{5}{4} \ln |x-3| + \frac{3}{4} \ln |x+1| \bigg|_{1}^{3}. \] Calculating the definite integral: At \( x = 3 \): \[ \frac{5}{4} \ln |3-3| + \frac{3}{4} \ln |3+1| = \text{undefined} + \frac{3}{4} \ln 4. \] At \( x = 1 \): \[ \frac{5}{4} \ln |1-3| + \frac{3}{4} \ln |1+1| = \frac{5}{4} \ln 2 + \frac{3}{4} \ln 2 = \frac{8}{4} \ln 2 = 2 \ln 2. \] As we see, the upper limit computes to infinity, thus suggesting the integral diverges specifically at \( x = 3 \). Therefore, the integral diverging cannot be computed within standard bounds: \[ \int_{1}^{3} \frac{2 x-1}{x^{2}-2 x-3} \, dx = \infty. \]