1. **State the problem:** We need to find the integral $$\int \frac{x^3 + 5}{x^2 - 25} \, dx$$. 2. **Rewrite the denominator:** Note that $$x^2 - 25 = (x - 5)(x + 5)$$. 3. **Perform polynomial division:** Since the degree of the numerator (3) is greater than the denominator (2), divide $$x^3 + 5$$ by $$x^2 - 25$$. Divide $$x^3$$ by $$x^2$$ to get $$x$$. Multiply $$x$$ by $$x^2 - 25$$ to get $$x^3 - 25x$$. Subtract: $$ (x^3 + 5) - (x^3 - 25x) = 25x + 5$$. So, $$\frac{x^3 + 5}{x^2 - 25} = x + \frac{25x + 5}{x^2 - 25}$$. 4. **Rewrite the integral:** $$\int \frac{x^3 + 5}{x^2 - 25} \, dx = \int x \, dx + \int \frac{25x + 5}{(x - 5)(x + 5)} \, dx$$. 5. **Partial fraction decomposition:** Set $$\frac{25x + 5}{(x - 5)(x + 5)} = \frac{A}{x - 5} + \frac{B}{x + 5}$$. Multiply both sides by $$ (x - 5)(x + 5) $$: $$25x + 5 = A(x + 5) + B(x - 5)$$. 6. **Find A and B:** Let $$x = 5$$: $$25(5) + 5 = A(5 + 5) + B(0) \Rightarrow 125 + 5 = 10A \Rightarrow 130 = 10A \Rightarrow A = 13$$. Let $$x = -5$$: $$25(-5) + 5 = A(0) + B(-5 - 5) \Rightarrow -125 + 5 = -10B \Rightarrow -120 = -10B \Rightarrow B = 12$$. 7. **Rewrite the integral:** $$\int x \, dx + \int \frac{13}{x - 5} \, dx + \int \frac{12}{x + 5} \, dx$$. 8. **Integrate each term:** $$\int x \, dx = \frac{x^2}{2} + C_1$$ $$\int \frac{13}{x - 5} \, dx = 13 \ln|x - 5| + C_2$$ $$\int \frac{12}{x + 5} \, dx = 12 \ln|x + 5| + C_3$$ 9. **Combine all results:** $$\int \frac{x^3 + 5}{x^2 - 25} \, dx = \frac{x^2}{2} + 13 \ln|x - 5| + 12 \ln|x + 5| + C$$ where $$C = C_1 + C_2 + C_3$$ is the constant of integration.