Question

Integrate by using the substitution suggested in bracket
$\int 12\left(y^4+4y^2+1\right)\left(y^3+2y\right)dy,$ , $\left(use,u=y^4+4y^2+1\right)$

Answer 1

$u=y^4+4y^2+1⟹du=(4y^3+8y)dy=4(y^3+2y)dy$

$\int 12(y^4+4y^2+1)(y^3+2y)dy=3\int udu=\frac{3}{2}{u}^2+c=\frac{3}{2}(y^4+2y^2+1{)}^2+c$