\( \int _{}^{}(3y^{2}+6)(y^{3}+3y^{2}+1)^{\frac{2}{3}} d y \)

To solve the integral \( \int (3y^{2}+6)(y^{3}+3y^{2}+1)^{\frac{2}{3}} \, dy \), we can use a substitution method. Let's define \( u = y^{3} + 3y^{2} + 1 \). Then, the derivative \( du = (3y^{2} + 6) \, dy \). Now, rewrite the integral in terms of \( u \): \[ dy = \frac{du}{3y^{2} + 6} \] Substituting this into the integral gives: \[ \int (3y^{2}+6)(u)^{\frac{2}{3}} \frac{du}{3y^{2}+6} = \int u^{\frac{2}{3}} \, du \] This integral is straightforward to evaluate: \[ \int u^{\frac{2}{3}} \, du = \frac{u^{\frac{5}{3}}}{\frac{5}{3}} + C = \frac{3}{5} u^{\frac{5}{3}} + C \] Finally, substituting back for \( u \): \[ = \frac{3}{5} (y^{3} + 3y^{2} + 1)^{\frac{5}{3}} + C \] So the solution to the integral is: \[ \int (3y^{2}+6)(y^{3}+3y^{2}+1)^{\frac{2}{3}} dy = \frac{3}{5} (y^{3} + 3y^{2} + 1)^{\frac{5}{3}} + C \] which provides the final answer!