How to integrate 1/(1+e^x) from (0

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

How to integrate 1/(1+e^x) from (0

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Try u substitution. What would you guess? What would you let u be?
u = 1+e^6?
Good guess, go for it.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Do you mean 1+e^x?
yes, thats what I meant, sorry
Aaronip, that actually won't work. This problem is actually pretty difficult. I can't think of anything to say but try u = e^x because that will work. However, you have to figure out how to use u in that way. Then you need to use partial fractions.
Yeah, this a special case.Let u be x. so integrate du/(1+e^u). It turns out this is equal to (-e^-u du)/((e^-u)+1). I don't know why. I have to research this, getting answer out of a book. Integrated it becomes ln |1+e^-u|. Evaluate over interval; there is a change of interval with the x to u thing. Good luck.
Okay, I'll try that. Thank you so much~!
Actually, it you can do it as u = e^x and then integrate du/(u*(1+u)) and then do partial fractions.But there might be an easier way.
aaronip, there's an easier way. Notice that\[\frac{1}{1+e^x}=\frac{1+e^x-e^x}{1+e^x}=\frac{1+e^x}{1+e^x}-\frac{e^x}{1+e^x}=1-\frac{e^x}{1+e^x}\]Then you'd have\[\int\limits_{0}^{1}\frac{dx}{1+e^x}=\int\limits_{0}^{1}1-\frac{e^x}{1+e^x}dx=\left[ x-\log (1+e^x) \right]_0^1\]\[=(1-\log (1+e))-(0-\log 2)=1+\log 2-\log (1+e)\]
Lokisan~ You're amazing~! Thank you~!
Good job lokisan I would like to see the machinations that make 1/(1+e^x) equal to (-e^-u du)/((e^-u)+1) as I found in a book.
haha, thank you :) Umm, what's your book thinking?
It's in the answer section, no explanation.
Are you sure the + sign in the denominator should be there?
Yeah, it says +
I get\[-\frac{e^{-u}}{e^{-u}-1}\] if I make the substitution,\[e^u = 1+e^x\]
I'm looking at it. to see what you did. One minute
I don't get it. what did you do?
1 Attachment
^^
Got it. So may the book distributed the negative sign at the bottom?
Maybe...that's assuming my guess for the book's substitution is correct.
Good job brilliant. Where did you learn these little tricks?
advanced degree in mathematics
Hoo, hoo. One day I will be up there.
Good! I'm off. Enjoy your book ;p

Not the answer you are looking for?

Search for more explanations.

Ask your own question