<s>[INST] How to finish this integration? 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 $dA=4x^3\,dx$
 [/INST] 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...
$$ = \frac{3}{4} \frac{A^{2/3}}{2/3} +C = \cdots$$
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