find the interval e^x (1+e^x)^2 dx

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- Feb 6th 2006, 09:19 PM #1

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- Feb 6th 2006, 09:48 PM #2

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- Feb 7th 2006, 02:04 PM #3

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Originally Posted by**kdstalnaker**

$\displaystyle \int e^x(1+e^x)^2dx$.

Okay, call $\displaystyle u=1+e^x$, then by substitution,

$\displaystyle \int u^2du=\frac{1}{3}u^3+C$

Thus,

$\displaystyle \frac{1}{3}e^{3x}+C$

Q.E.D.

- Feb 7th 2006, 08:19 PM #4

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Originally Posted by**ThePerfectHacker**

not the same. Then look at my posting; its a hint. Differentiate what

I suggest and you will recover the integrand. So what does that tell

you about the integral?

RonL

RonL

- Feb 8th 2006, 06:23 PM #5

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- Feb 8th 2006, 08:47 PM #6

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Originally Posted by**ThePerfectHacker**

$\displaystyle

\frac{d}{dx}\left(\frac{1}{3}e^{3x}+C \right)=e^{3x} \ne e^x(1+e^x)^2

$

and so $\displaystyle \frac{1}{3}e^{3x}+C$ is not the integral of $\displaystyle e^x(1+e^x)^2$

But:

$\displaystyle

\frac{d}{dx}\frac{(1+e^x)^3}{3}\ = e^x(1+e^x)^2

$

and so $\displaystyle \frac{(1+e^x)^3}{3}+C$ is the integral of $\displaystyle e^x(1+e^x)^2$

RonL

- Feb 9th 2006, 12:39 PM #7

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