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The given integral is:

\[
\int \frac{dx}{\sqrt{36x^2 - 1}}
\]

The hint suggests using inverse hyperbolic functions. Specifically, the form of the integrand is typical for an inverse hyperbolic secant or inverse hyperbolic cosine. Let’s proceed with solving the integral.

### Step-by-Step Solution:

1. **Rewrite the denominator**:
   \[
   \sqrt{36x^2 - 1} = \sqrt{(6x)^2 - 1}
   \]
   This is of the form \(\sqrt{a^2x^2 - b^2}\), which indicates we can apply a standard formula involving hyperbolic functions.

2. **Use the standard form**:
   The integral \(\int \frac{dx}{\sqrt{a^2x^2 - b^2}}\) has the antiderivative:
   \[
   \frac{1}{a} \text{arcosh} \left( \frac{ax}{b} \right) + C
   \]
   where \( \text{arcosh}(z) \) is the inverse hyperbolic cosine function.

   In our case:
   - \( a = 6 \)
   - \( b = 1 \)

   Therefore, applying the formula:
   \[
   \int \frac{dx}{\sqrt{36x^2 - 1}} = \frac{1}{6} \text{arcosh}(6x) + C
   \]

### Final Answer:
\[
\int \frac{dx}{\sqrt{36x^2 - 1}} = \frac{1}{6} \text{arcosh}(6x) + C
\]

Would you like more details on any step? Here are 5 related questions that might interest you:

1. How do you derive the inverse hyperbolic cosine formula for integrals?
2. Can we solve this integral using trigonometric substitution instead of hyperbolic functions?
3. What is the relationship between inverse hyperbolic functions and logarithms?
4. How can you verify the solution by differentiating the antiderivative?
5. What are some real-world applications of inverse hyperbolic functions in physics and engineering?

**Tip**: Always verify the correctness of an antiderivative by differentiating it, which should return the original integrand.

Find the antiderivative for the given function: ∫ dx / sqrt(36x^2 - 1).

Learn how to solve the integral ∫ dx / sqrt(36x^2 - 1) using inverse hyperbolic functions. This solution involves the formula for the inverse hyperbolic cosine (arcosh). Suitable for calculus students.

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