Answer
$\int 12(y^4 + 4y^2 + 1)^2(y^3+2y)dy = (y^4+4y^2+1)^3+C$
Work Step by Step
$\int 12(y^4 + 4y^2 + 1)^2(y^3+2y)dy\space$ and $\space u=y^4 + 4y^2 + 1$
$du = (4y^3 + 8y)dy$
Doing the substitution $\space u = y^4 + 4y^2 + 1$:
$\int(u)^2\cdot 3du\space\Rightarrow\space 3\int u^2 du$
Applying the integrative rules:
$3\int u^2 du = 3\cdot\frac{u^3}{3}+C\space\Rightarrow\space 3\int u^2 du = u^3+C$
Backing to $y$
$\int 12(y^4 + 4y^2 + 1)^2(y^3+2y)dy = (y^4+4y^2+1)^3+C$
