Ta có: $\int \dfrac{dx}{(3x-1)^2}$

$=\dfrac{1}{3} \int \dfrac{d(3x-1)}{(3x-1)^2}$

$=\dfrac{-1}{3}.\dfrac{1}{3x-1}+C$

$=\dfrac{-1}{9x-3}+C$

Vậy nguyên hàm của $\int \dfrac{dx}{(3x-1)^2}$ là $\dfrac{-1}{9x-3}+C$