Use partial fractions to find the indefinite integral. \[ \int \frac{x^{2}+108 x+108}{x^{3}-4 x} d x \]

To solve the integral \(\int \frac{x^{2}+108 x+108}{x^{3}-4 x} \, dx\), we first factor the denominator: \[ x^3 - 4x = x(x^2 - 4) = x(x - 2)(x + 2) \] Next, we set up the partial fraction decomposition: \[ \frac{x^{2}+108 x+108}{x^3 - 4x} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2} \] Multiplying both sides by the denominator \(x(x - 2)(x + 2)\) gives: \[ x^{2}+108 x+108 = A(x - 2)(x + 2) + Bx(x + 2) + Cx(x - 2) \] Expanding the right side: \[ A(x^2 - 4) + B(x^2 + 2x) + C(x^2 - 2x) = (A + B + C)x^2 + (2B - 2C)x + (-4A) \] Now we set the coefficients equal: 1. For \(x^2\) terms: \(A + B + C = 1\) 2. For \(x\) terms: \(2B - 2C = 108\) 3. For the constant terms: \(-4A = 108\) From the third equation: \[ A = -\frac{108}{4} = -27 \] Substituting \(A = -27\) into the first equation: \[ -27 + B + C = 1 \implies B + C = 28 \] Now, substituting \(A = -27\) into the second equation: \[ 2B - 2C = 108 \implies B - C = 54 \] Now we have a system of equations: 1. \(B + C = 28\) 2. \(B - C = 54\) Adding these two equations: \[ 2B = 82 \implies B = 41 \] Substituting \(B\) back into \(B + C = 28\): \[ 41 + C = 28 \implies C = -13 \] Now we have: \[ A = -27, \quad B = 41, \quad C = -13 \] Therefore, the partial fraction decomposition is: \[ \frac{x^{2}+108 x+108}{x^{3}-4 x} = \frac{-27}{x} + \frac{41}{x - 2} + \frac{-13}{x + 2} \] Now we can integrate term by term: \[ \int \frac{-27}{x} \, dx + \int \frac{41}{x - 2} \, dx + \int \frac{-13}{x + 2} \, dx \] Calculating each integral: 1. \(\int \frac{-27}{x} \, dx = -27 \ln |x| + C\) 2. \(\int \frac{41}{x - 2} \, dx = 41 \ln |x - 2| + C\) 3. \(\int \frac{-13}{x + 2} \, dx = -13 \ln |x + 2| + C\) Combining these results, we get: \[ \int \frac{x^{2}+108 x+108}{x^{3}-4 x} \, dx = -27 \ln |x| + 41 \ln |x - 2| - 13 \ln |x + 2| + C \] where \(C\) is the constant of integration.