The integral given is:

$\int \frac{x - 2}{(2x - 3)^2} \, dx$

To solve this integral, we can use substitution. Let's proceed with the following steps:

1. Substitute: Let $u = 2x - 3$. Then, the derivative $\frac{du}{dx} = 2$, or $dx = \frac{du}{2}$.

2. Rewrite $x - 2$ in terms of $u$: Since $u = 2x - 3$, we can solve for $x$ as follows: $x = \frac{u + 3}{2}$ Therefore, $x - 2 = \frac{u + 3}{2} - 2 = \frac{u + 3 - 4}{2} = \frac{u - 1}{2}$

3. Rewrite the integral in terms of $u$: Substitute $x - 2 = \frac{u - 1}{2}$ and $dx = \frac{du}{2}$: $\int \frac{x - 2}{(2x - 3)^2} \, dx = \int \frac{\frac{u - 1}{2}}{u^2} \cdot \frac{du}{2}$ Simplifying, this becomes:

   = \frac{1}{4} \int \left( \frac{u}{u^2} - \frac{1}{u^2} \right) du$$ $$= \frac{1}{4} \int \left( \frac{1}{u} - \frac{1}{u^2} \right) du$$

4. Integrate: Now, we can integrate each term separately: $= \frac{1}{4} \left( \ln |u| + \frac{1}{u} \right) + C$

5. Substitute back $u = 2x - 3$: $= \frac{1}{4} \left( \ln |2x - 3| + \frac{1}{2x - 3} \right) + C$

So, the solution is:

$\int \frac{x - 2}{(2x - 3)^2} \, dx = \frac{1}{4} \ln |2x - 3| + \frac{1}{4(2x - 3)} + C$

Would you like further details or have any questions?

Here are five related questions for further exploration:

1. How would this integral change if the denominator was $(2x - 3)^3$ instead of $(2x - 3)^2$?
2. Can we solve this integral using integration by parts instead of substitution?
3. How does the solution differ if we integrate $\frac{x + 2}{(2x - 3)^2}$?
4. What if we integrate $\frac{x - 2}{(2x - 3)^n}$ for an arbitrary power $n$?
5. How would we approach the integral if the numerator were a constant instead of $x - 2$?

Tip: When faced with rational functions in integrals, substitution is often useful, especially when the denominator’s structure can be simplified by redefining variables.