$\int y^{2}(y+2)^{\frac{1}{4}} d y$

Step 1 ：First, we recognize this as an integral of the form $\int u dv$, where $u = y^2$ and $dv = (y+2)^{\frac{1}{4}} dy$.

Step 2 ：We then need to find $du$ and $v$. We find $du$ by differentiating $u$ with respect to $y$, and we find $v$ by integrating $dv$ with respect to $y$.

Step 3 ：Differentiating $u = y^2$ with respect to $y$ gives $du = 2y dy$.

Step 4 ：Integrating $dv = (y+2)^{\frac{1}{4}} dy$ with respect to $y$ gives $v = \frac{4}{5}(y+2)^{\frac{5}{4}}$.

Step 5 ：We can now use the formula for integration by parts, which is $\int u dv = uv - \int v du$.

Step 6 ：Substituting $u = y^2$, $v = \frac{4}{5}(y+2)^{\frac{5}{4}}$, $du = 2y dy$, we get $\int y^{2}(y+2)^{\frac{1}{4}} dy = y^2 \cdot \frac{4}{5}(y+2)^{\frac{5}{4}} - \int \frac{4}{5}(y+2)^{\frac{5}{4}} \cdot 2y dy$.

Step 7 ：Simplify the integral to get $\frac{4}{5}y^2(y+2)^{\frac{5}{4}} - \frac{8}{5} \int y(y+2)^{\frac{5}{4}} dy$.

Step 8 ：We can see that the integral on the right side is still a bit complicated. We can use the substitution method to simplify it. Let $t = y+2$, then $dt = dy$, and when $y = t-2$, the integral becomes $\frac{8}{5} \int (t-2)t^{\frac{5}{4}} dt$.

Step 9 ：Expand the integral to get $\frac{8}{5} \int (t^{\frac{9}{4}} - 2t^{\frac{5}{4}}) dt$.

Step 10 ：Integrate to get $\frac{8}{5} (\frac{4}{13}t^{\frac{13}{4}} - \frac{4}{9}t^{\frac{9}{4}})$.

Step 11 ：Substitute $t = y+2$ back into the equation to get $\frac{8}{5} (\frac{4}{13}(y+2)^{\frac{13}{4}} - \frac{4}{9}(y+2)^{\frac{9}{4}})$.

Step 12 ：Substitute this result back into the equation we got from the integration by parts formula to get the final result: $\frac{4}{5}y^2(y+2)^{\frac{5}{4}} - \frac{8}{5} (\frac{4}{13}(y+2)^{\frac{13}{4}} - \frac{4}{9}(y+2)^{\frac{9}{4}})$.

Step 13 ：Simplify to get the final result: $\boxed{\frac{4}{5}y^2(y+2)^{\frac{5}{4}} - \frac{32}{65}(y+2)^{\frac{13}{4}} + \frac{32}{45}(y+2)^{\frac{9}{4}}}$.