To solve the integral

$\int \frac{6}{x^2 - 9} \, dx,$

we can first notice that the denominator can be factored:

$x^2 - 9 = (x - 3)(x + 3).$

Now we will split the fraction into partial fractions. The general form of partial fraction decomposition for this fraction is:

$\frac{6}{(x - 3)(x + 3)} = \frac{A}{x - 3} + \frac{B}{x + 3},$

where $A$ and $B$ are constants to be determined.

Step 1: Find $A$ and $B$

We multiply both sides of the equation by $(x - 3)(x + 3)$ to eliminate the denominators:

$6 = A(x + 3) + B(x - 3).$

Now, expand both sides:

$6 = A(x) + 3A + B(x) - 3B.$

Simplifying:

$6 = (A + B)x + (3A - 3B).$

For the equation to hold true for all values of $x$, the coefficients of $x$ and the constant terms must be equal on both sides. This gives us the system of equations:

1. $A + B = 0$
2. $3A - 3B = 6$

Step 2: Solve for $A$ and $B$

From the first equation, $A + B = 0$, we have $B = -A$.

Substitute $B = -A$ into the second equation:

$3A - 3(-A) = 6,$ $3A + 3A = 6,$ $6A = 6,$ $A = 1.$

Since $B = -A$, we have $B = -1$.

Step 3: Substitute $A$ and $B$ into the partial fractions

Now that we know $A = 1$ and $B = -1$, we can rewrite the integral as:

$\int \frac{6}{(x - 3)(x + 3)} \, dx = \int \left( \frac{1}{x - 3} - \frac{1}{x + 3} \right) \, dx.$

Step 4: Integrate

Now, integrate each term separately:

$\int \frac{1}{x - 3} \, dx = \ln |x - 3| + C_1,$ $\int \frac{1}{x + 3} \, dx = \ln |x + 3| + C_2.$

Thus, the solution to the integral is:

$\ln |x - 3| - \ln |x + 3| + C,$

where $C = C_1 - C_2$ is the constant of integration.

Final Answer:

$\int \frac{6}{x^2 - 9} \, dx = \ln \left| \frac{x - 3}{x + 3} \right| + C.$

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Would you like more details on any step, or have any further questions?

Here are some related questions you might find interesting:

1. What is the process of partial fraction decomposition?
2. How do you integrate rational functions involving quadratic expressions?
3. What is the meaning of the absolute value in logarithmic integrals?
4. How would the integral change if the denominator were a perfect square, like $x^2 - 4$?
5. How can partial fractions help with solving integrals that involve trigonometric functions?

Tip: Always check if the denominator can be factored before trying partial fraction decomposition!