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?
 A: 
$$\int{x^2(8x^3+27)^{2/3}}dx$$

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:
$$\dfrac 1{24} \dfrac {u^{5/3}}{5/3} + C = \dfrac {3}{24 \cdot 5}(8x^3 + 27)^{5/3} + C = \dfrac {1}{40}(8x^3 + 27)^{5/3} + C$$
A: Now, $u = 8x^3 + 27$ is a choice that makes less work for you.  But, one thing that students should understand is even if you don't make the best choice, it still might work.  You tried $u = 8x^3$.  In that case $du = 24x^2 \,dx$ and your integral becomes
$$\int{x^2(8x^3+27)^{2/3}}dx = \frac{1}{24} \int (u+27)^{2/3} \,du.$$
You might think to yourself, well, I can't do anything with that.  But, I am telling you that you can.  What if that was your original integral?  Might you not try a substitution of $v = u + 27$ with $dv = du$?  That would give you
$$\frac{1}{24} \int v^{2/3} \,dv = \frac{1}{24} \cdot \frac{3}{5} v^{5/3} + C.$$
At this point, you can undo both substitutions.
This method, in the end, is basically the same thing as doing $u = 8x^3 + 27$ in the first place but it's okay to make a choice that's not the best in the first place.  Just try to keep going from there.  If you can make the integral simpler at each step, then you are getting closer to solving the problem.
A: Let $u=8x^{3}+27$ then $du=24x^{2}dx$. So
$\displaystyle\int x^{2}(8x^{3}+27)^{\frac{2}{3}}dx=\frac{1}{24}\int(8x^{3}+27)^{\frac{2}{3}}(24x^{2})dx=\frac 1{24}\int u^{\frac{2}{3}}du$
A: You would use $u,du$ substitution
$$\int{x^2(8x^3+27)^{2/3}}dx$$
$$=\int{x^2(u)^{2/3}}dx$$
$$u=8x^3+27, du=24x^2 \implies \frac{du}{24}=x^2dx$$
$$\int{x^2(u)^{2/3}}dx$$
$$=\int{(u)^{2/3}x^2}dx$$
$$=\int{(u)^{2/3}}\frac{du}{24}$$
$$=\frac{1}{24}\int{(u)^{2/3}}du$$
$$=\frac{1}{24}\int{(u)^{2/3}}du$$
$$=\frac{1}{24}\left(\frac{(u)^{2/3+1}}{2/3+1}\right)+C$$
$$=\left(\frac{(u)^{5/3}}{24(5/3)}\right)+C$$
$$=\frac{u^{5/3}}{40}+C$$
