Answer
$$\frac{1}{2}\ln \left| {3{x^2} + 2x - 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{3x + 1}}{{3{x^2} + 2x - 1}}} dx \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{Let }}u = 3{x^2} + 2x - 1,\,\,\,du = \left( {6x + 2} \right)dx,\,\,\,\,\,\,\,dx = \frac{{du}}{{2\left( {3x + 1} \right)}} \cr
& {\text{write in terms of }}u \cr
& \int {\frac{{3x + 1}}{{3{x^2} + 2x - 1}}} dx = \int {\frac{{3x + 1}}{u}} \left( {\frac{{du}}{{2\left( {3x + 1} \right)}}} \right) \cr
& = \frac{1}{2}\int {\frac{{du}}{u}} \cr
& = \frac{1}{2}\ln \left| u \right| + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{2}\ln \left| {3{x^2} + 2x - 1} \right| + C \cr} $$
