Integrals, trouble with substitution problem

Question

This is my work on the problem, not sure if I did this wrong or I'm missing some way to simplify this and continue from here.

Answer 1

Let $u=6x^2-8x+3$. Then $(3x-2)\,dx$ is $\frac{du}{4}$. So you end up calculating $\int \frac{u^3}{4}\,du$.

Answer 2

Let $u=6x^2-8x+3$. Then $du=(12x-8)dx$=$4(3x-2)dx$. Then $\frac{du}{4}$=$(3x-2)dx$ So substituting $u$ and $du$ you get$\int \frac{u^3}{4}\,du$=$\frac{u^4}{16}$+$C$=$\frac{(6x^2-8x+3)^4}{16}$+$C$ Just wrote it all. No credit, as the answer was given above.