Evaluate the Integral: $\int\ x^2\ e^{x^3}\ dx$ $\int\ x^2\ e^{x^3}\ dx$
Step one: eliminate the $x$'s from the problem. The only way this can be done is utilizing $u$-substitution $u=x^3$ 
$u=x^3$
$du=3x^2\ dx$
$3\int\ e^u\ du$
at this point I am confused. Can some one point me in the correct direction? 
 A: Hint:
$\int e^u du$ means "the function(s) (of $u$) whose derivative is $e^u$".  Which function has a derivative of $e^u$?
A: if $\color{green}{u} = \color{magenta}{x^3}$ and $du = 3x^2 dx$, then $\color{red}{\frac{1}{3}du} = \color{blue}{x^2 dx}$, so
$$
\int x^2 e^{x^3}\;dx = \int e^{\color{magenta}{x^3}}\color{blue}{x^2 dx}
= \int e^{\color{green}{u}}\color{red}{\frac{1}{3}du}
$$
This integral is (without the colors)
$$
\frac{1}{3}\int e^u du = \frac{1}{3}e^u + C = \dots
$$
A: Your very last step is incorrect. You need to substitute $\frac{1}{3}{du} = {x^2 dx}$. Then you get $\frac{1}{3}\int e^u du$. The integral of this is $\frac{1}{3}(e^u) + c$, so the answer would be $\frac{1}{3}(e^{x^3}) + c$
A: I would like to start with a step zero: what is the derivative of $e^{x^3}$? 
Applying the chain rule we find: $3x^2e^{x^3}$. 
That quite "looks like" $x^2e^{x^3}$. 
Hmm, how to get rid of that constant $3$?

by multiplying with $\frac13$ so the answer is $\frac13e^{x^3}+C$

Step zero makes step one redundant.
