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.

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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:

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

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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 ]

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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:

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

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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:

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

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Final Answer

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

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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!