Answer
$\frac{1}{10}(2y+3)^{5/2}-\frac{1}{2} (2y+3)^{3/2}+c.$
Work Step by Step
Let $u=2y+3$, $du=2dy$. We have,
\begin{align*}
\int y(2y+3)^{1/2}dy&=\frac{1}{4}\int (u-3)u^{1/2}du\\
&=\frac{1}{4}\int u^{3/2}-3u^{1/2}du\\
&=\frac{1}{4}\frac{1}{5/2}u^{5/2}-\frac{1}{4}\frac{3}{3/2}u^{3/2}+c\\
&=\frac{1}{10}(2y+3)^{5/2}-\frac{1}{2} (2y+3)^{3/2}+c.
\end{align*}
