Another indefinite integral problem (e^x)

[irishpump]

Here's the problem: e^x/1+e^x dx

I let 1+e^x=U -> e^x/U -> e^x - u^-1 -> e^x/x - U^-1/-1, so does that mean x's and -1's would cancel? Wouldn't that make it lnu or ln(1+e^x) + C? I have no idea if I'm doing this right

 

[pka]

Here's the problem: e^x/1+e^x dx

What is the derivative of \(\displaystyle \ln(1+e^x)~?\)

 

[irishpump]

What is the derivative of \(\displaystyle \ln(1+e^x)~?\)

Wouldn't it be 1/1+e^x?

 

[srmichael]

Wouldn't it be 1/1+e^x?

Nope. If \(\displaystyle y=\ln(u)\) and u is a function of x, then \(\displaystyle \frac{dy}{du}=\frac{u\prime}{u}\). Try again!

 

[irishpump]

Nope. If \(\displaystyle y=\ln(u)\) and u is a function of x, then \(\displaystyle \frac{dy}{du}=\frac{u\prime}{u}\). Try again!

Okay, so if u'/u, then 1 is 0 because it's a constant, and e^x is still e^x, then it would be lnu -> ln(1+e^x) +c, which is e^x/1+e^x Thank you guys! I think the only thing that I find so frustrating about calculus is myself. I seem to always make things more complicated then they actually are, which causes me to get totally confused. Thanks again very much!

 

[lookagain]

Here's the problem: \(\displaystyle > > \) e^x/1+e^x \(\displaystyle \ < < \)dx

irishpump,

you have to stop typing this. It is not correct. You need grouping symbols,
because of the Order of Operations.


It is typed as:

e^x/(1 + e^x).


What you have is equivalent to (e^x)/1 + e^x = e^x + e^x = 2e^x.

 

[irishpump]

irishpump,

you have to stop typing this. It is not correct. You need grouping symbols,
because of the Order of Operations.


It is typed as:

e^x/(1 + e^x).


What you have is equivalent to (e^x)/1 + e^x = e^x + e^x = 2e^x.

Sorry