Let's solve the integral step by step: Given: \[ \int \frac{6x^3 + 5x^2 - 147x - 130}{x^2 - 25} \, dx \] ### 1. **Polynomial Division** The numerator is degree 3, and the denominator is degree 2, so perform polynomial division: #### Divide \(6x^3 + 5x^2 - 147x - 130\) by \(x^2 - 25\): - First term: \(6x^3 / x^2 = 6x\) - Multiply: \(6x \cdot (x^2 - 25) = 6x^3 - 150x\) - Subtract: \[ (6x^3 + 5x^2 - 147x - 130) - (6x^3 - 150x) = 5x^2 + 3x - 130 \] - Next term: \(5x^2 / x^2 = 5\) - Multiply: \(5 \cdot (x^2 - 25) = 5x^2 - 125\) - Subtract: \[ (5x^2 + 3x - 130) - (5x^2 - 125) = 3x - 5 \] So, \[ \frac{6x^3 + 5x^2 - 147x - 130}{x^2 - 25} = 6x + 5 + \frac{3x - 5}{x^2 - 25} \] ### 2. **Partial Fraction Decomposition** Factor denominator: \(x^2 - 25 = (x - 5)(x + 5)\) So, \[ \frac{3x - 5}{x^2 - 25} = \frac{A}{x - 5} + \frac{B}{x + 5} \] Multiply both sides by \((x - 5)(x + 5)\): \[ 3x - 5 = A(x + 5) + B(x - 5) \] Expand: \[ 3x - 5 = (A + B)x + (5A - 5B) \] Equate coefficients: - \(x\) terms: \(A + B = 3\) - Constant terms: \(5A - 5B = -5\) Solve system: - \(A + B = 3\) - \(A - B = -1\) (Divide the second equation by 5) Add equations: - \(A + B = 3\) - \(A - B = -1\) ---------------------- \(2A = 2 \implies A = 1\) Substitute \(A = 1\): - \(1 + B = 3 \implies B = 2\) ### 3. **Collect All Results** So, the decomposition is: \[ 6x + 5 + \frac{1}{x - 5} + \frac{2}{x + 5} \] ### 4. **Integrate Term by Term** \[ \int (6x + 5) \, dx + \int \frac{1}{x - 5} \, dx + \int \frac{2}{x + 5} \, dx \] Integrate: - \(\int 6x \, dx = 3x^2\) - \(\int 5 \, dx = 5x\) - \(\int \frac{1}{x - 5} \, dx = \ln|x - 5|\) - \(\int \frac{2}{x + 5} \, dx = 2\ln|x + 5|\) ### 5. **Final Answers** #### Values: - \(a = 6\) - \(b = 5\) - \(c = 1\) - \(d = 2\) #### The Final Integral: \[ \boxed{ \int \frac{6x^3 + 5x^2 - 147x - 130}{x^2 - 25} \, dx = 3x^2 + 5x + \ln|x - 5| + 2\ln|x + 5| + C } \]