• Feb 6th 2006, 09:19 PM
kdstalnaker
find the interval e^x (1+e^x)^2 dx
• Feb 6th 2006, 09:48 PM
CaptainBlack
Quote:

Originally Posted by kdstalnaker
find the interval e^x (1+e^x)^2 dx

What is the derivative of:

$\displaystyle \frac{(1+e^x)^3}{3}\ ?$

RonL
• Feb 7th 2006, 02:04 PM
ThePerfectHacker
Quote:

Originally Posted by kdstalnaker
find the interval e^x (1+e^x)^2 dx

I understand it as,
$\displaystyle \int e^x(1+e^x)^2dx$.
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
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
I understand it as,
$\displaystyle \int e^x(1+e^x)^2dx$.
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.

Try differentiating this, then compare it with the integrand. They are
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

RonL

RonL
• Feb 8th 2006, 06:23 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
Try differentiating this, then compare it with the integrand. They are
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

RonL

RonL

Not sure where you are getting at?
• Feb 8th 2006, 08:47 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Not sure where you are getting at?

What I am getting at is that

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

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
ThePerfectHacker
Of course! Look at how I used substitution rule. I subsituted $\displaystyle e^x$ for $\displaystyle u$ instead of $\displaystyle 1+e^x$.