Question

Find \[ \int \frac{x^2 + 1}{(x - 1)^2 (x + 3)} \, dx. \]

Hint:

To integrate rational functions, use partial fraction decomposition to break the function into simpler fractions and then integrate each term separately.

Answer 1

We are asked to evaluate the integral: \[ I = \int \frac{x^2 + 1}{(x - 1)^2 (x + 3)} \, dx. \] To solve this, we will use partial fraction decomposition. First, express the integrand as a sum of simpler fractions: \[ \frac{x^2 + 1}{(x - 1)^2 (x + 3)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{x + 3}. \] Multiply both sides by the denominator \((x - 1)^2 (x + 3)\): \[ x^2 + 1 = A(x - 1)(x + 3) + B(x + 3) + C(x - 1)^2. \] Now expand both sides: \[ x^2 + 1 = A(x^2 + 2x - 3) + B(x + 3) + C(x^2 - 2x + 1). \] Simplify: \[ x^2 + 1 = A(x^2 + 2x - 3) + B(x + 3) + C(x^2 - 2x + 1). \] Collect terms in powers of \(x\): \[ x^2 + 1 = (A + C)x^2 + (2A - 2C + B)x + (-3A + 3B + C). \] Now equate the coefficients of like powers of \(x\): 1. Coefficient of \(x^2\): \(A + C = 1\) 2. Coefficient of \(x\): \(2A - 2C + B = 0\) 3. Constant term: \(-3A + 3B + C = 1\) Solve this system of equations: From \(A + C = 1\), we get \(C = 1 - A\). Substitute \(C = 1 - A\) into the second equation: \[ 2A - 2(1 - A) + B = 0 \quad \Rightarrow \quad 2A - 2 + 2A + B = 0 \quad \Rightarrow \quad 4A + B = 2 \quad \Rightarrow \quad B = 2 - 4A. \] Substitute \(C = 1 - A\) and \(B = 2 - 4A\) into the third equation: \[ -3A + 3(2 - 4A) + (1 - A) = 1 \quad \Rightarrow \quad -3A + 6 - 12A + 1 - A = 1 \quad \Rightarrow \quad -16A + 7 = 1 \quad \Rightarrow \quad -16A = -6 \quad \Rightarrow \quad A = \frac{3}{8}. \] Now substitute \(A = \frac{3}{8}\) into \(C = 1 - A\) and \(B = 2 - 4A\): \[ C = 1 - \frac{3}{8} = \frac{5}{8}, \quad B = 2 - 4 \times \frac{3}{8} = 2 - \frac{12}{8} = \frac{4}{8} = \frac{1}{2}. \] Thus, the partial fraction decomposition is: \[ \frac{x^2 + 1}{(x - 1)^2 (x + 3)} = \frac{\frac{3}{8}}{x - 1} + \frac{\frac{1}{2}}{(x - 1)^2} + \frac{\frac{5}{8}}{x + 3}. \] Now integrate each term: \[ I = \frac{3}{8} \int \frac{1}{x - 1} \, dx + \frac{1}{2} \int \frac{1}{(x - 1)^2} \, dx + \frac{5}{8} \int \frac{1}{x + 3} \, dx. \] The integrals are straightforward: \[ I = \frac{3}{8} \ln |x - 1| - \frac{1}{2(x - 1)} + \frac{5}{8} \ln |x + 3| + C. \]