To solve the given integral:

$\int \frac{dx}{36 + x^2}$

we'll use a standard formula for integrating a function of the form:

$\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C,$

where $a$ is a constant.

Here, we have $a^2 = 36$, so $a = 6$.

Applying the formula:

$\int \frac{dx}{36 + x^2} = \frac{1}{6} \tan^{-1}\left(\frac{x}{6}\right) + C.$

The correct answer is:

$\boxed{\text{C: } \frac{1}{6} \tan^{-1}\left(\frac{x}{6}\right) + C}$

Would you like more details or have any questions?

Related Questions:

1. How do you integrate functions of the form $\int \frac{dx}{a^2 + x^2}$?
2. What is the geometric interpretation of the arctangent function?
3. Can you solve an example where the integral is $\int \frac{dx}{25 + x^2}$?
4. What happens if $a$ is a function of $x$ instead of a constant in $\int \frac{dx}{a^2 + x^2}$?
5. How does the result change if we integrate $\int \frac{x \, dx}{36 + x^2}$?

Tip: Remember that integrals involving $a^2 + x^2$ often lead to inverse trigonometric functions like $\tan^{-1}$.