<s>[INST] Integrate $\int x^2(8x^3+27)^{2/3}\,dx$ Integrate $$\int{x^2(8x^3+27)^{2/3}}dx$$
I'm just wondering, what should I make $u$ equal to?
I tried to make $u=8x^3$, but it's not working. 
Can I see a detailed answer?
 [/INST] 
$$\int{x^2(8x^3+27)^{2/3}}dx$$

It is certainly possible to work with $u = 8x^3$, but I'd suggest setting $u = 8x^3 + 27$. The key is to remember to compute and account for $\,du$. What you'll see is that for both $u = 8x^3$ and $u = 8x^3 + 27$, we have $du = 24x^2$.   
$$u = 8x^3 + 27 \implies\,du = 24x^2 \,dx \iff  \color{blue}{\bf\dfrac 1{24} \,du = x^2 \,dx}$$
Substituting: $$\int x^2(8x^3+27)^{2/3}dx = \int (8x^3 + 27)^{2/3} \color{blue}{\bf x^2 \,dx} =   \color{blue}{\bf \dfrac1{24}} \int u^{2/3} \, \color{blue}{\bf du}$$
Evaluating the integral: $$\color{blue}{\bf \dfrac1{24}} \int u^{2/3} \, \color{blue}{\bf du} = \dfrac 1{24} \dfrac {u^{5/3}}{5/3} + C $$ 
Back substituting, given $u = 8x^3 + 27$, then simplifying, gives us:
$$\dfrac 1{24} \dfrac {u^{5/3}}{5/3} + C = \dfrac {3}{24 \cdot 5}(8x^3 + 27)^{5/3} + C = \dfrac {1}{40}(8x^3 + 27)^{5/3} + C$$
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