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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 $du=3x^2 dx$

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See this question –  Américo Tavares Jul 19 '11 at 22:28
    
It was asked the same integral and the same substitution. –  Américo Tavares Jul 19 '11 at 22:30
    
Indeed, the possible duplicate (math.stackexchange.com/questions/51983/…) –  mixedmath Jul 20 '11 at 1:56
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5 Answers 5

up vote 8 down vote accepted

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!

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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.

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We set $u=x^3$. Then, what is $\frac{d}{dx}x^3$? It is $3x^2$. That is why $du=3x^2 dx$.

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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 $n=3$.

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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 $$\begin{align*} \int x^2e^{x^3}dx &=\int 1\cdot x^2e^{x^3}dx\\ &=\int\frac{1}{3}\cdot 3x^2e^{x^3}dx\\ &=\frac{1}{3}\int 3x^2e^{x^3}dx=\frac{1}{3}\int g'(x)f(g(x))dx\\ &=\frac{1}{3}F(g(x))\\ &=\frac{1}{3}e^{x^3}. \end{align*}$$

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