$\displaystyle \int y^3\sqrt{2y^2 + 1} \,\, dy$
$\qquad = \displaystyle \int y^2 \cdot \sqrt{2y^2 + 1} \cdot y \, dy$
Let
$u = \sqrt{2y^2 + 1}$
$u^2 = 2y^2 + 1$
$y^2 = \frac{1}{2}(u^2 - 1)$
$2y \, dy = \frac{1}{2}(2u \, du)$
$y \, dy = \frac{1}{2}u \, du$
Hence,
$\displaystyle \int y^3\sqrt{2y^2 + 1} \,\, dy$
$\qquad = {\displaystyle \int} \frac{1}{2}(u^2 - 1) \cdot u \cdot \frac{1}{2}u \, du$
$\qquad = {\displaystyle \frac{1}{4}\int} (u^4 - u^2) \, du$
$\qquad = \dfrac{1}{4} \left( \dfrac{u^5}{5} - \dfrac{u^3}{3} \right) + C$
$\qquad = \dfrac{1}{4} \left( \dfrac{3u^5 - 5u^3}{15} \right) + C$
$\qquad = \frac{1}{60} u^3 (3u^2 - 5) + C$
Revert to original variable of integration
$\displaystyle \int y^3\sqrt{2y^2 + 1} \,\, dy$
$\qquad = \frac{1}{60} \left( \sqrt{2y^2 + 1} \right)^3 \left[ 3\left( \sqrt{2y^2 + 1} \right)^2 - 5 \right] + C$
$\qquad = \frac{1}{60} (2y^2 + 1)^{3/2} [ \, 3(2y^2 + 1) - 5 \, ] + C$
$\qquad = \frac{1}{60} (2y^2 + 1)^{3/2} [ \, 6y^2 + 3 - 5 \, ] + C$
$\qquad = \frac{1}{60} (2y^2 + 1)^{3/2} (6y^2 - 2) + C$
$\qquad = \frac{1}{30} (2y^2 + 1)^{3/2} (3y^2 - 1) + C$ ← answer
