Consider the indefinite integral $\int \frac{x-3}{(3 x-2)^{2}} d x$. The substitution $u=3 x-2$ transforms the integral into:
$\int \frac{\frac{u-2}{3}-3}{u^{2}} d u$
None of these sptions are correct.
$\int \frac{u-\frac{7}{3}}{u^{2}} d u$
$\int \frac{1}{u^{2}} d u$
$\int \frac{u-7}{9 u^{2}} d u$

Step 1 ：Consider the indefinite integral \(\int \frac{x-3}{(3 x-2)^{2}} d x\). The substitution \(u=3 x-2\) transforms the integral into: \(\int \frac{\frac{u-2}{3}-3}{u^{2}} d u\).

Step 2 ：None of these options are correct: \(\int \frac{u-\frac{7}{3}}{u^{2}} d u\), \(\int \frac{1}{u^{2}} d u\), \(\int \frac{u-7}{9 u^{2}} d u\).

Step 3 ：The integral \(\int \frac{x-3}{(3 x-2)^{2}} d x\) can be rewritten as \(\int \frac{x-3}{u^{2}} d x\) after the substitution \(u=3x-2\). We also know that \(du/dx = 3\), so \(dx = du/3\). Substituting these into the integral, we get \(\int \frac{x-3}{u^{2}} \cdot \frac{du}{3}\).

Step 4 ：We can simplify this further by substituting \(x\) in terms of \(u\), which gives \(x = (u+2)/3\). Let's substitute this into the integral and simplify.

Step 5 ：Substituting \(x = u/3 + 2/3\) into the integral, we get \(\int \frac{u/3 + 2/3 -3}{u^{2}} \cdot \frac{du}{3}\).

Step 6 ：The integral simplifies to \(\frac{1}{3} \log(u) + \frac{7}{3u}\). This is not in the form of any of the given options.

Step 7 ：Final Answer: \(\boxed{\text{None of these options are correct}}\).