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

2. Originally Posted by kdstalnaker
find the interval e^x (1+e^x)^2 dx
What is the derivative of:

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

RonL

3. Originally Posted by kdstalnaker
find the interval e^x (1+e^x)^2 dx
I understand it as,
$\int e^x(1+e^x)^2dx$.
Okay, call $u=1+e^x$, then by substitution,
$\int u^2du=\frac{1}{3}u^3+C$
Thus,
$\frac{1}{3}e^{3x}+C$
Q.E.D.

4. Originally Posted by ThePerfectHacker
I understand it as,
$\int e^x(1+e^x)^2dx$.
Okay, call $u=1+e^x$, then by substitution,
$\int u^2du=\frac{1}{3}u^3+C$
Thus,
$\frac{1}{3}e^{3x}+C$
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

5. 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?

6. Originally Posted by ThePerfectHacker
Not sure where you are getting at?
What I am getting at is that

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

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

But:

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

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

RonL

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