Question

Find: \[ I = \int \frac{2x + 3}{x^2(x + 3)} \, dx \]

Hint:

When integrating rational functions, split the numerator and simplify into partial fractions if necessary to solve term by term.

Answer 1

Step 1: Split the integral. The given integral can be rewritten as: \[ I = \int \frac{x + 3}{x^2(x + 3)} \, dx + \int \frac{x}{x^2(x + 3)} \, dx \] Simplifying: \[ I = \int \frac{1}{x^2} \, dx + \int \frac{x + 3 - x}{x(x + 3)} \, dx \] Step 2: Further simplify. Split the second integral: \[ \int \frac{x + 3 - x}{x(x + 3)} \, dx = \int \frac{1}{x} \, dx - \int \frac{1}{x + 3} \, dx \] Step 3: Evaluate the individual integrals. 1. First term: \[ \int \frac{1}{x^2} \, dx = -\frac{1}{x} \] 2. Second term: \[ \int \frac{1}{x} \, dx = \log |x| \] 3. Third term: \[ \int \frac{1}{x + 3} \, dx = \log |x + 3| \] Step 4: Combine the results. Substituting back, we get: \[ I = -\frac{1}{x} + \frac{1}{3} \log |x| - \frac{1}{3} \log |x + 3| + C \] Conclusion: The final result is: \[ I = -\frac{1}{x} + \frac{1}{3} \log |x| - \frac{1}{3} \log |x + 3| + C \]