Answer
$$\frac{{4{{\left( {3x - 2} \right)}^{3/2}}}}{{27}} + \frac{{14{{\left( {3x - 2} \right)}^{1/2}}}}{9} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2x + 1}}{{\sqrt {3x - 2} }}} dx \cr
& {\text{Integrate by substitution method }}\left( {{\text{See note 1 in the book}}} \right) \cr
& {\text{ }}u = 3x - 2,{\text{ }}x = \frac{{u + 2}}{3},{\text{ }}dx = \frac{1}{3}du \cr
& {\text{Substituting}} \cr
& \int {\frac{{2x + 1}}{{\sqrt {3x - 2} }}dx} = \int {\frac{{2\left( {\frac{{u + 2}}{3}} \right) + 1}}{{\sqrt u }}\left( {\frac{1}{3}} \right)} du \cr
& \int {\frac{{\left( {2u/3 + 4/3 + 1} \right)}}{{{u^{1/2}}}}\left( {\frac{1}{3}} \right)} du \cr
& \int {\frac{{\left( {2u/3 + 7/3} \right)}}{{{u^{1/2}}}}\left( {\frac{1}{3}} \right)} du \cr
& \frac{1}{9}\int {\frac{{2u + 7}}{{{u^{1/2}}}}du} \cr
& \frac{1}{9}\int {\left( {\frac{{2u}}{{{u^{1/2}}}} + \frac{7}{{{u^{1/2}}}}} \right)du} \cr
& \frac{1}{9}\int {\left( {2{u^{1/2}} + 7{u^{ - 1/2}}} \right)du} \cr
& {\text{Integrate use the power rule }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C} \cr
& \frac{1}{9}\left[ {\frac{{2{u^{3/2}}}}{{3/2}} + \frac{{7{u^{1/2}}}}{{1/2}}} \right] + C \cr
& \frac{1}{9}\left[ {\frac{{4{u^{3/2}}}}{3} + 14{u^{1/2}}} \right] + C \cr
& \frac{{4{u^{3/2}}}}{{27}} + \frac{{14}}{9}{u^{1/2}} + C \cr
& {\text{Write in terms of }}x,{\text{ }}u = 3x - 2 \cr
& \frac{{4{{\left( {3x - 2} \right)}^{3/2}}}}{{27}} + \frac{{14{{\left( {3x - 2} \right)}^{1/2}}}}{9} + C \cr} $$
