Thank you for the clarification. The integral is:

$\int_{8}^{9} \frac{12}{x^2 - 36} \, dx$

Let's solve this integral using partial fraction decomposition and evaluate it over the correct limits.

Step 1: Factor the denominator

The denominator is factored as:

$x^2 - 36 = (x - 6)(x + 6)$

Step 2: Partial Fraction Decomposition

We express $\frac{12}{(x - 6)(x + 6)}$ as:

$\frac{12}{(x - 6)(x + 6)} = \frac{A}{x - 6} + \frac{B}{x + 6}$

Multiply both sides by $(x - 6)(x + 6)$:

$12 = A(x + 6) + B(x - 6)$

Expanding and simplifying:

$12 = A(x + 6) + B(x - 6) = Ax + 6A + Bx - 6B$

Combine like terms:

$12 = (A + B)x + (6A - 6B)$

This gives the system of equations:

1. $A + B = 0$
2. $6A - 6B = 12$

From the first equation, $A = -B$. Substituting into the second equation:

$6(-B) - 6B = 12$ $-6B - 6B = 12$ $-12B = 12 \quad \Rightarrow \quad B = -1$

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

Thus, the partial fraction decomposition is:

$\frac{12}{(x - 6)(x + 6)} = \frac{1}{x - 6} - \frac{1}{x + 6}$

Step 3: Evaluate the Integral

Now, evaluate the integral:

$\int_{8}^{9} \left(\frac{1}{x - 6} - \frac{1}{x + 6}\right) dx$

This can be separated as:

$\int_{8}^{9} \frac{1}{x - 6} \, dx - \int_{8}^{9} \frac{1}{x + 6} \, dx$

Each integral is of the form $\int \frac{1}{u} du = \ln |u|$, so:

$\int_{8}^{9} \frac{1}{x - 6} \, dx = \ln |x - 6| \Big|_{8}^{9} = \ln |9 - 6| - \ln |8 - 6| = \ln(3) - \ln(2)$

$\int_{8}^{9} \frac{1}{x + 6} \, dx = \ln |x + 6| \Big|_{8}^{9} = \ln |9 + 6| - \ln |8 + 6| = \ln(15) - \ln(14)$

So, the integral is:

$\left(\ln\left(\frac{3}{2}\right)\right) - \left(\ln\left(\frac{15}{14}\right)\right)$

This simplifies to:

$\ln\left(\frac{3}{2}\right) - \ln\left(\frac{15}{14}\right) = \ln\left(\frac{3}{2} \cdot \frac{14}{15}\right) = \ln\left(\frac{42}{30}\right) = \ln\left(\frac{7}{5}\right)$

Finally, compute this:

$\ln\left(\frac{7}{5}\right) \approx 0.336$

Conclusion

The correct answer is approximately $0.336$.

Would you like further details or have any other questions?

---

Related Questions

1. How does changing the limits of integration affect the result?
2. What are other examples where partial fractions are used to simplify integrals?
3. What would be the result if the integral was evaluated from 6 to 8 instead?
4. How would you approach this problem if the denominator was not factorable easily?
5. What is the general rule for setting up partial fraction decomposition for more complex denominators?

Tip:

Always simplify the expression inside the logarithm after integrating, as it can sometimes reduce to a simpler form.