We need to evaluate the integral:

\[
I = \int \frac{1}{2 - \sqrt{3x}} \, dx
\]

### Step 1: Substitution

Let’s use a substitution to simplify the square root expression. Set:

\[
u = \sqrt{3x}
\]

Then, differentiate \( u \) with respect to \( x \):

\[
du = \frac{1}{2\sqrt{3x}} \cdot 3 \, dx = \frac{3}{2\sqrt{3x}} \, dx \quad \Rightarrow \quad dx = \frac{2}{3} \, du
\]

Also, since \( u = \sqrt{3x} \), we have \( x = \frac{u^2}{3} \).

### Step 2: Rewrite the integral

Substitute these into the integral:

\[
I = \int \frac{1}{2 - u} \cdot \frac{2}{3} \, du
\]

Simplifying:

\[
I = \frac{2}{3} \int \frac{1}{2 - u} \, du
\]

### Step 3: Solve the integral

This is a standard integral of the form \( \int \frac{1}{a - u} \, du \), which equals \( -\ln |a - u| \). So:

\[
\int \frac{1}{2 - u} \, du = -\ln |2 - u|
\]

Thus, the integral becomes:

\[
I = \frac{2}{3} \cdot (-\ln |2 - u|) + C = -\frac{2}{3} \ln |2 - u| + C
\]

### Step 4: Back-substitute \( u = \sqrt{3x} \)

Finally, substitute \( u = \sqrt{3x} \) back into the expression:

\[
I = -\frac{2}{3} \ln |2 - \sqrt{3x}| + C
\]

### Final Answer:

\[
\boxed{I = -\frac{2}{3} \ln |2 - \sqrt{3x}| + C}
\]

Would you like any additional details or explanations?

Here are some related questions you can explore:
1. How do you handle integrals involving square roots in general?
2. What is a common substitution technique for integrals like this?
3. How would the result change if the integrand had a different constant inside the square root?
4. What is the significance of the absolute value in the logarithmic result?
5. Can this technique be applied to more complex functions inside the integral?

**Tip**: Always check if a substitution can simplify the expression involving roots or powers.

Learn how to evaluate the integral \( \int \frac{1}{2 - \sqrt{3x}} \, dx \) using the substitution method and logarithmic integration. This solution involves substitution with \( u = \sqrt{3x} \) and solving a standard form integral. Suitable for undergraduate calculus students.

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