What's wrong in my solution to $\int \frac{1}{(3x - 2)^2} dx$? I'm solving it this way:
$$u = 3x - 2\\
\frac{du}{3} = dx$$  
$$\frac{1}{3} \int \frac{1}{u^2} du \\
= \frac{1}{3} \frac{3}{u^3} + C$$

As David has pointed out, $$\int\frac{1}{u^2}\mathop{du}=-u^{-1}+C$$
