I'm working with the integral below, but not sure how to finish it...
$$\int \frac{3x^3}{\sqrt[3]{x^4+1}}\,dx = \int \frac{3x^3}{\sqrt[3]{A}}\cdot \frac{dA}{4x^3} = \frac{3}{4} \int \frac{dA}{\sqrt[3]{A}} = \frac{3}{4}\cdot\quad???$$ where $A=x^4+1$ and so $dA=4x^3\,dx$
Assuming I interpreted your very low quality picture correctly...
You got a good start on the problem. That's the right substitution. Rewrite...
$$\frac{3}{4} \int \frac{dA}{\sqrt[3]{A}} = \frac{3}{4} \int A^{-1/3}\,dA$$
...and use the backwards power rule...
$$ = \frac{3}{4} \frac{A^{2/3}}{2/3} +C = \cdots$$
$$\dfrac{1}{\sqrt[\large 3]{A}} = \dfrac 1{A^{1/3}} = A^{-1/3}$$
Now use the power rule.
$$ \frac{3}{4} \int A^{-1/3}\,dA = \frac 34 \dfrac {A^{2/3}}{\frac 23} + C = \dfrac 98 A^{2/3} + C$$
