The integral provided is:

$\int_{\sqrt{3}}^{3} \frac{dx}{x \sqrt{4x^2 - 9}}$

To solve this integral, we can use a substitution method. Let’s proceed step by step.

Step 1: Substitution

Let $u = 2x$, which gives $du = 2dx$ or $dx = \frac{du}{2}$.

Then we adjust the limits of integration accordingly:

• When $x = \sqrt{3}$, $u = 2\sqrt{3}$.
• When $x = 3$, $u = 6$.

The integral becomes:

$\frac{1}{2} \int_{2\sqrt{3}}^{6} \frac{du}{u \sqrt{u^2 - 9}}$

This is now in a standard form for the logarithmic integral:

$\int \frac{du}{u \sqrt{u^2 - a^2}} = \frac{1}{a} \ln \left| \frac{u + \sqrt{u^2 - a^2}}{a} \right| + C$

Here, $a = 3$, so we apply this formula.

Step 2: Applying the Formula

The integral becomes:

$\frac{1}{6} \ln \left| \frac{u + \sqrt{u^2 - 9}}{3} \right| \Bigg|_{2\sqrt{3}}^{6}$

Step 3: Evaluating the Limits

Now, we evaluate at the upper and lower limits.

For $u = 6$:

$\frac{6 + \sqrt{6^2 - 9}}{3} = \frac{6 + \sqrt{36 - 9}}{3} = \frac{6 + \sqrt{27}}{3} = \frac{6 + 3\sqrt{3}}{3} = 2 + \sqrt{3}$

For $u = 2\sqrt{3}$:

$\frac{2\sqrt{3} + \sqrt{(2\sqrt{3})^2 - 9}}{3} = \frac{2\sqrt{3} + \sqrt{12 - 9}}{3} = \frac{2\sqrt{3} + \sqrt{3}}{3} = \frac{\sqrt{3}(2 + 1)}{3} = \sqrt{3}$

Step 4: Final Expression

The integral result is:

$\frac{1}{6} \ln \left( \frac{2 + \sqrt{3}}{\sqrt{3}} \right)$

This is the final solution for the integral.

Would you like any further clarification or details on specific steps?

Here are five questions related to this integral:

1. What are the criteria for using the substitution method in integrals?
2. How does changing the limits of integration work after substitution?
3. Can you explain how to identify standard forms of integrals like the logarithmic one used here?
4. How would the result change if the limits of integration were different?
5. What other methods could be used to solve integrals like this one?

Tip: When solving integrals, always check if a substitution or standard form can simplify the problem significantly.