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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 $4x^{2}e^{u}dx$?

3 Answers 3

1

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. $

2

The mistake you have is that you should write $du = 3x^2 dx$ instead of $du = 3x^2$...

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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.2011-07-17
1

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.$2011-07-17