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I was a little stuck on this problem, so I took a look at the solution work and am confused as to how it works.

Given $\int{x^{2}e^{x^{3}}}dx$, find the indefinite integral.

Looking at the formula, I decided to have $u=x^{3}$ and $du=3x^{2}$

Multiplying and dividing by 3 on both sides, $$\int{x^{2}e^{x^{3}}}dx= \frac{1}{3}\int{x^{2}e^{u}(3x^{2})}dx$$

However, the book solution omits the $x^{2}$ in the second step, leaving just $\frac{1}{3}\int{e^{u}(3x^{2})}$dx

I have two questions:
1. why is the $x^{2}$ omitted in the book solution after multiplying and dividing by 3?
2. why does the solution not include simplifying $x^2e^{u}(3x^{2}) dx$ to result in $4x^{2}e^{u}dx$?

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3 Answers

up vote 1 down vote accepted

If $u=x^3$, then $\dfrac{du}{dx}=\dfrac{d}{dx}x^3=3x^2$ and $du=3x^2dx$, because the differential of $u=f(x)$ is $du=f'(x)dx=\dfrac{du}{dx}dx$.

EDIT: From $du=3x^{2}dx$, follows that $dx=\dfrac{du}{3x^{2}}$. Hence $$\int x^{2}e^{x^{3}}dx=\int x^{2}e^{u}\frac{du}{3x^{2}}=\int \frac{1}{3}e^{u}\,du.$$

EDIT 2: Thus $$\int x^{2}e^{x^{3}}dx=\int \frac{1}{3}e^{u}\,du=\frac{1}{3}e^{u}+C=\frac{1}{3}e^{x^{3}}+C.$$

EDIT 3: Integration by substitution technique. If we change variables by making the substitution $u=f(x)$, then we have

$$\int g(x)dx=\int g(f(u))f^{\prime }(u)du=\int g(u)\frac{dx}{du}du,$$

and

$$dx=f^{\prime }(u)du=\frac{dx}{du}du.$$

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Note: One could have computed $dx$ using the inverse rule for derivatives (and differentials). $$dx=\dfrac{1}{\frac{du}{dx}}du.$$ –  Américo Tavares Jul 17 '11 at 18:05
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The mistake you have is that you should write $du = 3x^2 dx$ instead of $du = 3x^2$...

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But isn't that dx more semantical than anything else? –  Jason Jul 17 '11 at 17:41
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@Jason: No. The differential of $u=f(x)$ is $du=f'(x)dx$. –  Américo Tavares Jul 17 '11 at 17:43
    
@Jason: I suggest you go over the statement of the substitution rule and try to see how your steps fit in, and where you are mistaken. dx tells you the variable you are integrating with respect to. What exactly do you mean by semantical? –  Aryabhata Jul 17 '11 at 18:00
2  
@Jason: André's comment here might interest you... –  J. M. Jul 17 '11 at 18:08
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@JM, I think I fit both parts of Andre's comment, as almost all the integration I've done is with respect to x, so I find I omit the $\frac{d}{dx}$ quite often. Maybe thats a habit I should break. –  Jason Jul 17 '11 at 18:35
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Using $u=x^3$ and, hence, $du=3x^2 \, dx$, you get $$ \int {x^2 e^{x^3 } dx} = \frac{1}{3}\int {e^{x^3 } 3x^2 \,dx} = \frac{1}{3}\int {e^u \,du} . $$

EDIT: It follows that $$ \int {x^2 e^{x^3 } dx} = \frac{1}{3} e^{x^3} + C. $$

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