Answer
$$\int12(y^4+4y^2+1)^2(y^3+2y)dy=(y^4+4y^2+1)^3+C$$
Work Step by Step
$$A=\int12(y^4+4y^2+1)^2(y^3+2y)dy$$
We set $u=y^4+4y^2+1$.
Then $$du=(4y^3+8y)dy=4(y^3+2y)dy$$
That means $$(y^3+2y)dy=\frac{1}{4}du$$
Therefore, $$A=\int\Big(12u^2\times\frac{1}{4}du\Big)=\int3u^2du$$ $$A=\frac{3u^3}{3}+C=u^3+C$$ $$A=(y^4+4y^2+1)^3+C$$