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Integration by Substitution

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
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!
Reply 2
I don't do partial fractions till C4, this is a C3 question (AQA)
Reply 3
well, you get

du = e^x dx

which is du = (u -1) dx.

or -(u-1) du = dx.

_Kar.

Hint:

Spoiler

(edited 13 years ago)
Reply 4
Yes that is incorrect, thanks for trying
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.
Reply 6
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
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).
Reply 8
I agree with the latter!
Reply 9
about the error in the book, I doubt you're missing something!
Reply 10
We've I=2ex+1  dxI = \int\frac{2}{e^x+1}\;{dx}. Letting u=ex+1u = e^x+1 gives 2u(u1)  du\int\frac{2}{u(u-1)}\;{du} . You don't have to know the method of partial fractions to observe that 2=2u2(u1)2 = 2u-2(u-1).
Thus I=2u2(u1)u(u1)  du=21u1  du21u  duI= \int\frac{2u-2(u-1)}{u(u-1)}\;{du} = 2\int\frac{1}{u-1}\;{du}-2\int\frac{1}{u}\;{du} =2ln(u1)2lnu+k=2lnex2ln(ex+1)+k=2xln(ex+1)+k.= 2\ln\left(u-1\right)-2\ln{u}+k = 2\ln{e^x}-2\ln\left(e^x+1\right)+k = 2x-\ln\left(e^x+1\right)+k.
(edited 13 years ago)

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