Question about indefinite integral Let's say I have expression with multiplication which has variable x
$$\int x^2e^{x^3}dx$$
So in example it shows 
$$\int x^2e^{x^3}dx=\frac{1}{3}\int e^u du=\frac{1}{3}e^u+C=\frac{e^{x^3}}{3}+C$$
$u=x^3$ $du=3x^2 dx$
So I don't understand from where comes $\frac{1}{3}$ before integral and from where comes 3 in $du=3x^2 dx$ 
 A: This is called substitution.
Here are the steps in detail:
You want to make two replacements:  $u=x^3$ and $du=3x^2 dx$. But you don't have $3x^2 dx$ in your integral. No matter, construct it! Starting with what you have:
$$\int x^2e^{x^3}dx $$
multiply by 1=3/3:
$$= \frac{3}{3}\int x^2e^{x^3}dx $$
move the 3 inside of the integral and move $x^2$ next to the $dx$:
$$= \frac{1}{3}\int e^{x^3}3x^2dx$$
Then make two replacements:  $u=x^3$ and $du=3x^2 dx$.
$$ =\frac{1}{3}\int e^u du$$
These replacements are compatible with each other because if you differentiate $u=x^3$ you get $du=3x^2 dx$. In general, if what you are looking to substitute ($3x^2 dx$ in this case) differs only by a constant from what you have ($x^2 dx$) , you can introduce the constant you need by putting 1 over that constant "outside" the integral. 
Your best bet is to do another very similar example right away so that this concept will solidify for you. Best!
A: Substituting, you have
$$
\int x^2e^{x^3}dx=\int x^2e^{u}\frac{du}{3x^2}=\frac{1}{3}\int e^u du
$$
after canceling the $x^2$. The $3$ in $du=3x^2 dx$ comes from the fact that the derivative of $x^3$ is $3x^2$ by the power rule.
A: We set $u=x^3$.  Then, what is $\frac{d}{dx}x^3$?  It is $3x^2$.  That is why $du=3x^2 dx$.
A: If you have some integral of the form $$\int x^{n-1} e^{x^n} \ dx$$ the best thing to do is to put $t = x^{n}$. What then happens is $dt =nx^{n-1} dx$. Now you have $x^{n-1}\cdot dx$ inside the integral and you want to express everything in terms of $t$. You see that $\frac{dt}{n} = x^{n-1} \ dx$. So put this inside and your integral becomes $$\int e^{t} \cdot \frac{1}{n} \cdot dt$$ which is precisely what you have done for the case $n=3$.
A: The substitutions are made to simplify things. I think that you should understand why it works.
If $F$ is primitive of $f$ (i.e. $F'(x)=f(x)$) and $g$ is a function such that $f(g(x))$ and $F(g(x))$, both, makes sense, then the function $$f(g(x))g'(x)$$ has a primitive, namely $F(g(x))$. This follows by the chain rule.
In your case
$$f(t)= e^t,\qquad F(t)=e^t,\qquad\text{and}\qquad g(x)=x^3,$$ thus $$f(g(x))=e^{x^3}$$ and you want $g'(x)f(g(x))$. Now, $$g'(x)=3x^2$$ and you already have $x^2$, so you must produce $3x^2$ in some way. In this case it is easy because
$$\begin{align*}
\int x^2e^{x^3}dx &=\int 1\cdot x^2e^{x^3}dx\\
                  &=\int\frac{1}{3}\cdot 3x^2e^{x^3}dx\\
                  &=\frac{1}{3}\int 3x^2e^{x^3}dx=\frac{1}{3}\int g'(x)f(g(x))dx\\
                  &=\frac{1}{3}F(g(x))\\
                  &=\frac{1}{3}e^{x^3}.
\end{align*}$$
