The given integral is:

\[
\int \frac{3x^2 - 6x - 27}{x^3 - 9x} \, dx
\]

### Step 1: Factor the denominator and numerator
We can factor both the numerator and the denominator.

- Denominator: \( x^3 - 9x = x(x^2 - 9) = x(x - 3)(x + 3) \)
- Numerator: \( 3x^2 - 6x - 27 = 3(x^2 - 2x - 9) \)

So the integral becomes:

\[
\int \frac{3(x^2 - 2x - 9)}{x(x - 3)(x + 3)} \, dx
\]

### Step 2: Partial Fraction Decomposition
We express the integrand as partial fractions:

\[
\frac{3(x^2 - 2x - 9)}{x(x - 3)(x + 3)} = \frac{A}{x} + \frac{B}{x - 3} + \frac{C}{x + 3}
\]

Multiplying both sides by the denominator \( x(x - 3)(x + 3) \), we get:

\[
3(x^2 - 2x - 9) = A(x - 3)(x + 3) + Bx(x + 3) + Cx(x - 3)
\]

### Step 3: Solve for A, B, and C
Now, expand and compare coefficients on both sides:

- \( A(x^2 - 9) + B(x^2 + 3x) + C(x^2 - 3x) \)

Equating the coefficients:

- Coefficient of \(x^2\): \( 3 = A + B + C \)
- Coefficient of \(x\): \( -6 = 3B - 3C \)
- Constant term: \( -27 = -9A \)

From the constant term:  
\[
A = 3
\]

Substitute \( A = 3 \) into the other equations:

1. \( 3 = 3 + B + C \Rightarrow B + C = 0 \)
2. \( -6 = 3B - 3C \Rightarrow B - C = -2 \)

From \( B + C = 0 \) and \( B - C = -2 \), solving these two gives:
- \( B = -1 \)
- \( C = 1 \)

### Step 4: Rewrite the integral
Now that we have \( A = 3 \), \( B = -1 \), and \( C = 1 \), the integral becomes:

\[
\int \left( \frac{3}{x} - \frac{1}{x - 3} + \frac{1}{x + 3} \right) dx
\]

### Step 5: Integrate
Now, integrate each term:

\[
3 \int \frac{1}{x} \, dx - \int \frac{1}{x - 3} \, dx + \int \frac{1}{x + 3} \, dx
\]

The integrals are straightforward:

\[
3 \ln |x| - \ln |x - 3| + \ln |x + 3| + C
\]

### Step 6: Combine logarithmic terms
Using logarithmic properties:

\[
3 \ln |x| - \ln |x - 3| + \ln |x + 3| = \ln \left( \frac{x^3(x + 3)}{x - 3} \right)
\]

Thus, the final answer is:

\[
\ln \left( \frac{x^3(x + 3)}{x - 3} \right) + C
\]

Would you like any clarifications or further details?

### 5 Related Questions:
1. How do you solve partial fractions for higher-degree polynomials?
2. What is the role of logarithmic properties in simplifying integrals?
3. Can we apply the same method if the degree of the numerator is higher than the denominator?
4. What happens if the polynomial in the denominator has complex roots?
5. How can we integrate rational functions involving irreducible quadratic factors?

**Tip:** When solving integrals of rational functions, always check if partial fraction decomposition is possible—it simplifies the process considerably.

Learn how to solve the integral of (3x^2 - 6x - 27)/(x^3 - 9x) using partial fraction decomposition and logarithmic properties. This calculus problem involves factoring polynomials, decomposing fractions, and applying logarithmic integration rules. Suitable for college-level or advanced high school calculus students.

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