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Math help? D:

Compute the indefinite integral of 1/(1+e^t)

Let's go calc students

March 24, 2011

6 Comments • Newest first

Obscene

Learn math pls. Joanna.
S[1/(1+e^t)]dt
=S[(1+e^t-e^t)/(1+e^t)]dt
=S[1]dt - S[e^t / (1+e^t)]dt
=t - S[e^t / (1+e^t)]dt
Let u = 1+e^t
du = e^t * dt
There for,
S[1/(1+e^t)]dt = t - S[1 / u]du
= t - ln|u| + C
= t - ln|1+e^t| + C

Reply March 24, 2011
blippster

The problem with metaghost's answer is that he is integrating a (dt). He has to convert dt to da by finding the derivative of a where a=some function of t (1+e^t). THen you solve for integral of a(da). But it's almost impossible to solve it this way without using integration by tables which is really complicated..

Reply March 24, 2011 - edited
Splutter

[quote=samlee95]Differentiate t - ln(1+e^t) gives:
1 - e^t/(1+e^t) = 1/(1+e^t)
---
Differentiate e^t * ln(1+e^t) gives:
e^t * e^t/(1+e^t) + e^t * ln(1+e^t)
I think it is because 1/a * a' dt = ln(a) + C, not dt/a = ln(a) * a' + C[/quote]

My hero *sparkly eyes*

thank you

Reply March 24, 2011 - edited
Splutter

so which is it:

t - ln(1+e^t) + C or e^t * ln(1+e^t) + C...

Reply March 24, 2011 - edited
Splutter

[quote=finkle]e^t (ln 1+e^t) + C[/quote]
I was hoping more of an explanation *_* how?

Reply March 24, 2011 - edited
MatthewDough

I copypasta'd your question into wolframalpha.
http://www.wolframalpha.com/input/?i=Compute+the+indefinite+integral+of+1%2F%281%2Be^t%29

So I have no idea what happened.

Reply March 24, 2011 - edited