The given problem asks you to evaluate the improper integral:

$\int_6^{\infty} \frac{dx}{x^2 - 25}$

This is an improper integral due to the infinite upper limit.

Steps to solve:

1. Factor the denominator: $x^2 - 25 = (x - 5)(x + 5)$ The integral becomes: $\int_6^{\infty} \frac{dx}{(x - 5)(x + 5)}$

2. Use partial fraction decomposition: We express the integrand as a sum of partial fractions: $\frac{1}{(x - 5)(x + 5)} = \frac{A}{x - 5} + \frac{B}{x + 5}$ To find $A$ and $B$, solve the equation: $1 = A(x + 5) + B(x - 5)$ Expanding and combining like terms: $1 = (A + B)x + 5A - 5B$ Set up the system of equations:

   • $A + B = 0$
   • $5A - 5B = 1$

   Solving this gives:

   • $A = \frac{1}{10}$
   • $B = -\frac{1}{10}$

   So, the integral becomes: $\int_6^{\infty} \left( \frac{1}{10(x - 5)} - \frac{1}{10(x + 5)} \right) dx$

3. Integrate each term: $\frac{1}{10} \int_6^{\infty} \left( \frac{1}{x - 5} - \frac{1}{x + 5} \right) dx$ The integral of $\frac{1}{x - 5}$ is $\ln|x - 5|$, and the integral of $\frac{1}{x + 5}$ is $\ln|x + 5|$.

   So, we get: $\frac{1}{10} \left[ \ln|x - 5| - \ln|x + 5| \right]_6^{\infty}$

4. Evaluate the limits:

   • As $x \to \infty$, both $\ln|x - 5|$ and $\ln|x + 5|$ tend to infinity, but they cancel out, giving 0.
   • At $x = 6$: [ \frac{1}{10} \left[ \ln(6 - 5) - \ln(6 + 5) \right] = \frac{1}{10} \left[ \ln 1 - \ln 11 \right] = \frac{1}{10}(- \ln 11) ]

   So, the value of the integral is: $-\frac{1}{10} \ln 11$

Conclusion:

The improper integral evaluates to $-\frac{1}{10} \ln 11$, which corresponds to the answer $\boxed{\frac{1}{10} \ln 11}$ in the provided options.

Would you like further details or explanations?

Related Questions:

1. How does partial fraction decomposition work for higher-degree polynomials?
2. Why does an improper integral with an infinite limit require special handling?
3. What happens if the integrand had other singularities within the limits of integration?
4. Can you explain the importance of convergence tests for improper integrals?
5. How does the natural logarithm function behave as $x$ approaches infinity?

Tip:

Always check if the improper integral converges before attempting to evaluate it—this avoids unnecessary calculations.