To solve the integral

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

we will use the method of substitution. Let us set

$u = y + 2$

then

$du = dy$

and since

$y = u - 2,$

we can rewrite the integral in terms of $u$:

$\int y^2 (y + 2)^{\frac{1}{4}} \, dy = \int (u - 2)^2 u^{\frac{1}{4}} \, du.$

Now, expand $(u - 2)^2$:

$(u - 2)^2 = u^2 - 4u + 4.$

Substituting this back into the integral, we get:

$\int (u^2 - 4u + 4) u^{\frac{1}{4}} \, du.$

Now, distribute $u^{\frac{1}{4}}$:

$\int u^2 \cdot u^{\frac{1}{4}} - 4u \cdot u^{\frac{1}{4}} + 4 \cdot u^{\frac{1}{4}} \, du.$

This simplifies to:

$\int u^{2 + \frac{1}{4}} - 4u^{1 + \frac{1}{4}} + 4u^{\frac{1}{4}} \, du = \int u^{\frac{9}{4}} - 4u^{\frac{5}{4}} + 4u^{\frac{1}{4}} \, du.$

Integrate each term separately:

$\int u^{\frac{9}{4}} \, du = \frac{u^{\frac{9}{4} + 1}}{\frac{9}{4} + 1} = \frac{u^{\frac{13}{4}}}{\frac{13}{4}} = \frac{4}{13} u^{\frac{13}{4}},$

$\int -4u^{\frac{5}{4}} \, du = -4 \cdot \frac{u^{\frac{5}{4} + 1}}{\frac{5}{4} + 1} = -4 \cdot \frac{u^{\frac{9}{4}}}{\frac{9}{4}} = -\frac{16}{9} u^{\frac{9}{4}},$

$\int 4u^{\frac{1}{4}} \, du = 4 \cdot \frac{u^{\frac{1}{4} + 1}}{\frac{1}{4} + 1} = 4 \cdot \frac{u^{\frac{5}{4}}}{\frac{5}{4}} = \frac{16}{5} u^{\frac{5}{4}}.$

Combining these results, we have:

$\int y^2 (y + 2)^{\frac{1}{4}} \, dy = \frac{4}{13} u^{\frac{13}{4}} - \frac{16}{9} u^{\frac{9}{4}} + \frac{16}{5} u^{\frac{5}{4}} + C.$

Finally, substitute back $u = y + 2$:

$\frac{4}{13} (y + 2)^{\frac{13}{4}} - \frac{16}{9} (y + 2)^{\frac{9}{4}} + \frac{16}{5} (y + 2)^{\frac{5}{4}} + C.$

This is the solution to the integral. Do you have any questions or would you like more details?

Here are 5 questions you might ask:

1. Can you explain the method of substitution in integrals?
2. How do you choose the substitution $u = y + 2$ in this integral?
3. What are the steps to simplify the integral after substitution?
4. How do you integrate each term separately in the expanded integral?
5. Can you show another example of using substitution in integrals?

Tip: When performing substitution in integrals, always remember to change both the variable and the differential to match the substitution.