Answer
$$ - \frac{2}{{15}}\ln \left| {\frac{x}{{3x - 5}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{2}{{3x\left( {3x - 5} \right)}}} dx \cr
& {\text{drop out the constant }}\frac{2}{3} \cr
& = \frac{2}{3}\int {\frac{1}{{x\left( {3x - 5} \right)}}} dx \cr
& {\text{integrate by tables using the formulas on the apendix D for this book}} \cr
& {\text{using the formula 13}}:\,\,\,\,\,\,\,\int {\frac{1}{{x\left( {ax + b} \right)}}dx = \frac{1}{b} \cdot \ln \left| {\frac{x}{{ax + b}}} \right|} + C{\text{ }} \cr
& {\text{in the integral }}\int {\frac{1}{{x\left( {3x - 5} \right)}}} dx{\text{ we can see that }}a = 3{\text{ and }}b = - 5 \cr
& {\text{then}} \cr
& \frac{2}{3}\int {\frac{1}{{x\left( {3x - 5} \right)}}} dx = \frac{2}{3}\left( {\frac{1}{{ - 5}} \cdot \ln \left| {\frac{x}{{3x - 5}}} \right|} \right) + C \cr
& {\text{simplifying}} \cr
& = - \frac{2}{{15}}\ln \left| {\frac{x}{{3x - 5}}} \right| + C \cr} $$
