Notice that if you substitute $\displaystyle \begin{align*} u = \ln{ (3x - 5)} \end{align*}$ then $\displaystyle \begin{align*} \mathrm{d}u = \frac{3}{3x - 5}\,\mathrm{d}x \end{align*}$, so write your integral as

$\displaystyle \begin{align*} \int{ \frac{12\ln{(3x-5)}}{3x-5}\,\mathrm{d}x} &= 4\int{ \ln{(3x-5)}\,\frac{3}{3x-5}\,\mathrm{d}x } \\ &= 4\int{ u\,\mathrm{d}u} \\ &= 4 \left( \frac{u^2}{2} \right) + C \\ &= 2u^2 + C \\ &= 2\left[ \ln{(3x-5)} \right] ^2 + C \end{align*}$