The integral you provided is:

\[
\int_0^1 \frac{1}{x^2 - 9} \, dx
\]

This integrand contains a rational function with a quadratic denominator, and it looks like it could benefit from **partial fraction decomposition** or recognizing the form directly. However, before proceeding, note that \(x^2 - 9\) has **real roots** at \(x = \pm 3\). These roots are outside the interval of integration \([0, 1]\), so there are no singularities within this range, and the integral is well-defined.

---

### Step 1: Decomposing the integrand using partial fractions

We start by expressing the integrand \( \frac{1}{x^2 - 9} \) as a sum of simpler rational functions. Notice that:

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

So, we decompose:

\[
\frac{1}{x^2 - 9} = \frac{A}{x - 3} + \frac{B}{x + 3}
\]

Multiplying both sides by \(x^2 - 9 = (x - 3)(x + 3)\) yields:

\[
1 = A(x + 3) + B(x - 3)
\]

Expanding and rearranging terms:

\[
1 = (A + B) \cdot x + (3A - 3B)
\]

This equation holds for all \(x\), so we equate coefficients of \(x\) and the constant terms:

\[
A + B = 0 \quad \text{(1)}
\]
\[
3A - 3B = 1 \quad \text{(2)}
\]

From equation (1), \(B = -A\). Substituting into equation (2):

\[
3A - 3(-A) = 1 \quad \Rightarrow \quad 6A = 1 \quad \Rightarrow \quad A = \frac{1}{6}
\]

Thus, \(B = -\frac{1}{6}\). So the partial fraction decomposition is:

\[
\frac{1}{x^2 - 9} = \frac{1/6}{x - 3} - \frac{1/6}{x + 3}
\]

---

### Step 2: Integrating the decomposed form

Now, we integrate:

\[
\int_0^1 \frac{1/6}{x - 3} \, dx - \int_0^1 \frac{1/6}{x + 3} \, dx
\]

Both integrals are of the form \( \int \frac{1}{x - a} \, dx = \ln |x - a| \), so:

\[
\frac{1}{6} \left[ \ln |x - 3| \right]_0^1 - \frac{1}{6} \left[ \ln |x + 3| \right]_0^1
\]

---

### Step 3: Evaluating the definite integrals

1. **For** \( \ln |x - 3| \) from 0 to 1:

   \[
   \left[ \ln |x - 3| \right]_0^1 = \ln |1 - 3| - \ln |0 - 3| = \ln 2 - \ln 3 = \ln \left(\frac{2}{3}\right)
   \]

2. **For** \( \ln |x + 3| \) from 0 to 1:

   \[
   \left[ \ln |x + 3| \right]_0^1 = \ln |1 + 3| - \ln |0 + 3| = \ln 4 - \ln 3 = \ln \left(\frac{4}{3}\right)
   \]

---

### Step 4: Simplifying the result

Now, substitute the results into the integral:

\[
\frac{1}{6} \left( \ln \left(\frac{2}{3}\right) \right) - \frac{1}{6} \left( \ln \left(\frac{4}{3}\right) \right)
\]

Using properties of logarithms:

\[
\ln \left(\frac{2}{3}\right) - \ln \left(\frac{4}{3}\right) = \ln \left(\frac{2/3}{4/3}\right) = \ln \left(\frac{2}{4}\right) = \ln \left(\frac{1}{2}\right) = -\ln 2
\]

Thus, the integral becomes:

\[
\frac{1}{6} \cdot (-\ln 2) = -\frac{\ln 2}{6}
\]

---

### Final Answer

\[
\int_0^1 \frac{1}{x^2 - 9} \, dx = -\frac{\ln 2}{6}
\]

---

Let me know if you need further clarifications or steps!  
Here are 5 related questions you might find interesting:

1. How do you handle integrals with quadratic denominators that have complex roots?
2. What is the partial fraction decomposition technique used for higher-degree polynomials?
3. How does the improper integral of \( \frac{1}{x^2 - 9} \) behave over an infinite interval?
4. Can you solve similar integrals involving \(x^2 + 9\) instead of \(x^2 - 9\)?
5. How does the natural logarithm appear in solutions to integrals involving rational functions?

**Tip:** When integrating rational functions, always check if partial fractions can simplify the problem!

This page provides a step-by-step solution to the definite integral \int_0^1 \frac{1}{x^2 - 9} dx, utilizing partial fraction decomposition and logarithmic integration. Explore key calculus concepts such as partial fractions and properties of definite integrals, with detailed explanations suitable for undergraduate students.

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