We have $\lim_{n\to\infty}\ln{(\frac{\sqrt[n+1]{(...)}}{n+1})}=\lim_{n\to\infty}\frac{1}{n+1}(\sum_{i=1}^{n+1}\ln{(2+\frac{i}{n+1})})=\int_{0}^{1}\ln{(2+x)}dx.$

\[\lim_{n\to \infty}\frac{\sqrt[n+1]{(2n+3)(2n+4)\ldots (3n+3)}}{n+1}=2\lim_{n\to \infty}exp(\frac{1}{n+1}\sum_{k=1}^{n+1}ln(1+\frac{k}{2n+2}))=2exp(\int_{0}^{1}ln(1+\frac{x}{2})dx)=\frac{9}{2e}.\]

The other classical method: If $a_{n}= \frac{\sqrt[n+1]{(2n+3)(2n+4)\ldots (3n+3)}}{n+1}$, then \[b_{n}= \ln a_{n}= \frac1{n+1}\cdot \left( \sum_{k=0}^{n}\ln (2n+3+k)-(n+1) \ln (n+1) \right) . \] By Cesaro-Stolz we get that \begin{eqnarray*}\lim_{n \to \infty}b_{n}&=& \lim_{n \to \infty}\left( \ln[(3n+4)(3n+5)(3n+6)]-\ln[(2n+3)(2n+4)(n+2)]+\ln \frac{(n+1)^{n+1}}{(n+2)^{n+1}}\right) \\ \ &=& \ln \frac{27}{4e},\end{eqnarray*} so $\lim_{n \to \infty}a_{n}= \frac{27}{4e}$. (Maybe I made some computational mistakes since I don't get the same answer.)