\int \frac{36}{\left(x^2-9\right)} partial fraction

To evaluate the integral \(\int \frac{36}{x^2 - 9} \, dx\) using partial fractions, follow these steps:

### Step 1: Factor the denominator
The denominator \(x^2 - 9\) is a difference of squares, so we can factor it as:
\[
x^2 - 9 = (x - 3)(x + 3)
\]

### Step 2: Set up the partial fraction decomposition
We express the integrand as a sum of fractions with unknown coefficients:
\[
\frac{36}{(x - 3)(x + 3)} = \frac{A}{x - 3} + \frac{B}{x + 3}
\]
Our goal is to find \(A\) and \(B\).

### Step 3: Solve for \(A\) and \(B\)
Multiply both sides by \((x - 3)(x + 3)\) to eliminate the denominators:
\[
36 = A(x + 3) + B(x - 3)
\]
Expand the right-hand side:
\[
36 = A(x + 3) + B(x - 3) = A x + 3A + B x - 3B
\]
Group the terms involving \(x\):
\[
36 = (A + B)x + (3A - 3B)
\]
Now, equate the coefficients of \(x\) and the constant terms:
- Coefficient of \(x\): \(A + B = 0\)
- Constant term: \(3A - 3B = 36\)

### Step 4: Solve the system of equations
From \(A + B = 0\), we get \(A = -B\).

Substitute \(A = -B\) into the second equation:
\[
3(-B) - 3B = 36 \quad \Rightarrow \quad -3B - 3B = 36 \quad \Rightarrow \quad -6B = 36 \quad \Rightarrow \quad B = -6
\]
Since \(A = -B\), we find that \(A = 6\).

### Step 5: Rewrite the integrand
Now that we have \(A = 6\) and \(B = -6\), we can rewrite the integrand:
\[
\frac{36}{(x - 3)(x + 3)} = \frac{6}{x - 3} - \frac{6}{x + 3}
\]

### Step 6: Integrate
Now integrate each term separately:
\[
\int \frac{6}{x - 3} \, dx - \int \frac{6}{x + 3} \, dx = 6 \ln |x - 3| - 6 \ln |x + 3| + C
\]

### Step 7: Combine the logarithms
Using the properties of logarithms, we can combine the terms:
\[
6 \ln |x - 3| - 6 \ln |x + 3| = 6 \ln \left|\frac{x - 3}{x + 3}\right|
\]

Thus, the final answer is:
\[
\int \frac{36}{x^2 - 9} \, dx = 6 \ln \left|\frac{x - 3}{x + 3}\right| + C
\]

Would you like more details or have any questions?

Here are 5 related questions for practice:
1. How would you solve \(\int \frac{2x}{x^2 - 9} \, dx\)?
2. Can you find the integral of \(\frac{36}{(x^2 + 9)}\)?
3. How would you decompose \(\frac{3x + 4}{(x-2)(x+3)}\) using partial fractions?
4. What is the integral of \(\frac{10}{x^2 - 16}\) using partial fractions?
5. How would you approach \(\int \frac{1}{x^2 - 1} \, dx\)?

**Tip**: When factoring quadratics, always check if it can be written as a difference of squares—it simplifies the partial fraction decomposition process.

Learn how to solve the integral ∫ 36/(x^2 - 9) using partial fraction decomposition. This detailed guide walks through factoring the difference of squares, setting up the decomposition, and solving the integral step by step. Suitable for advanced high school and college students.

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