Answer
$$ - \frac{1}{{\left( {4x + 6} \right)}} - \frac{1}{6}\ln \left| {\frac{x}{{4x + 6}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{ - 6}}{{x{{\left( {4x + 6} \right)}^2}}}} dx \cr
& {\text{drop out the constant }} - 6 \cr
& = - 6\int {\frac{1}{{x{{\left( {4x + 6} \right)}^2}}}} dx \cr
& {\text{integrate by tables using the formulas on the apendix D for this book}} \cr
& {\text{using the formula 14}}:\,\,\,\,\int {\frac{1}{{x{{\left( {ax + b} \right)}^2}}}dx = \frac{1}{{b\left( {ax + b} \right)}} + \frac{1}{{{b^2}}} \cdot \ln \left| {\frac{x}{{ax + b}}} \right|} + C{\text{ }}\left( {b \ne 0} \right) \cr
& {\text{in the integral }}\int {\frac{1}{{x{{\left( {4x + 6} \right)}^2}}}} dx{\text{ we can see that }}a = 4{\text{ and }}b = 6 \cr
& {\text{then}} \cr
& - 6\int {\frac{1}{{x{{\left( {4x + 6} \right)}^2}}}} dx = - 6\left( {\frac{1}{{6\left( {4x + 6} \right)}} + \frac{1}{{{6^2}}} \cdot \ln \left| {\frac{x}{{4x + 6}}} \right|} \right) + C \cr
& {\text{simplifying}} \cr
& - 6\int {\frac{1}{{x{{\left( {4x + 6} \right)}^2}}}} dx = - \frac{1}{{\left( {4x + 6} \right)}} - \frac{1}{6}\ln \left| {\frac{x}{{4x + 6}}} \right| + C \cr} $$
