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# Integration by Substitution watch

1. Could someone please give me a hint on how to do this integral?

2/(e^x+1) dx u=e^x+1

I get to integrating 2/(u^2-u) but stuck from here
2. (Original post by sdgvrgverg)
Could someone please give me a hint on how to do this integral?

2/(e^x+1) dx u=e^x+1

I get to integrating 2/(u^2-u) but stuck from here
Partial fractions!
3. I don't do partial fractions till C4, this is a C3 question (AQA)
4. well, you get

du = e^x dx

which is du = (u -1) dx.

or -(u-1) du = dx.

_Kar.

Hint:
Spoiler:
Show
The answer I (possibly incorrectly) reached was Ln (u) - 2u + C. That might help if you're looking for the method.
5. Yes that is incorrect, thanks for trying
6. (Original post by sdgvrgverg)
I don't do partial fractions till C4, this is a C3 question (AQA)
If you're going to use that substitution straight off, you're going to end up with partial fractions.

The only other thing I can suggest is to multiply top and bottom by e^(-x), and then use a similar substitution, which will get you round the partial fractions issue.
7. I'll try that, thanks.
Strange though, it's in the C3 section of the book and says to use the substitution u=e^x+1
8. (Original post by sdgvrgverg)
I'll try that, thanks.
Strange though, it's in the C3 section of the book and says to use the substitution u=e^x+1
It's possible I'm missing something, and also possible there's an error in the book (won't be the first time, in either case).
9. I agree with the latter!
10. about the error in the book, I doubt you're missing something!
11. We've . Letting gives . You don't have to know the method of partial fractions to observe that .
Thus

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