Answer
$$ \frac{{{{\left( {2y + 1} \right)}^{3/2}}}}{6} - \frac{{{{\left( {2y + 1} \right)}^{1/2}}}}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{y}{{\sqrt {2y + 1} }}} dy \cr
& {\text{substitute }}u = 2y + 1,{\text{ }}du = 2dy \cr
& \int {\frac{y}{{\sqrt {2y + 1} }}} dy = \int {\frac{{\left( {u - 1} \right)/2}}{{\sqrt u }}} \left( {\frac{1}{2}du} \right) \cr
& = \frac{1}{4}\int {\frac{{u - 1}}{{{u^{1/2}}}}} du \cr
& = \frac{1}{4}\int {\left( {{u^{1/2}} - {u^{ - 1/2}}} \right)} du \cr
& {\text{find antiderivative }} \cr
& = \frac{1}{4}\left( {\frac{{{u^{3/2}}}}{{3/2}} - \frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr
& = \frac{{{u^{3/2}}}}{6} - \frac{{{u^{1/2}}}}{2} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = 2y + 1 \cr
& = \frac{{{{\left( {2y + 1} \right)}^{3/2}}}}{6} - \frac{{{{\left( {2y + 1} \right)}^{1/2}}}}{2} + C \cr} $$
