Answer
$$\frac{4}{9}\left( {3x + \ln \left| {3x - 1} \right|} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{4x}}{{3x - 1}}} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{Rewrite the integrand}} \cr
& \int {\frac{{4x}}{{3x - 1}}} dx = 4\int {\frac{{xdx}}{{ - 1 + 3x}}} \cr
& {\text{The integrand has a expression in the form }}a + bu{}{\text{}} \cr
& {\text{Use the formula 60}} \cr
& \left( {60} \right):\,\,\,\,\,\int {\frac{{udu}}{{a + bu}}} = \frac{1}{{{b^2}}}\left[ {bu - a\ln \left| {a + bu} \right|} \right] + C \cr
& {\text{let }}u = x,\,\,\,a = - 1{\text{ and }}b = 3 \cr
& 4\int {\frac{{xdx}}{{ - 1 + 3x}}} = 4\left( {\frac{1}{{{{\left( 3 \right)}^2}}}\left[ {3x - \left( { - 1} \right)\ln \left| { - 1 + 3x} \right|} \right]} \right) + C \cr
& {\text{simplifying}} \cr
& \int {\frac{{4x}}{{3x - 1}}} dx = \frac{4}{9}\left( {3x + \ln \left| {3x - 1} \right|} \right) + C \cr} $$
