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# Integration help!! watch

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1. Hey guys, could anyone please explain how to do this question, i end up with +ln2 at the end so I can’t put it into a fraction:

Using the substitution u=e^x +1 show that the integral of e^(2x)/(e^x +1) with limits 1 to 0 is equal to e - 1 - ln((e+1)/2)

(Question 6 from the June 2006 OCR paper in case that’s easier to look at)

Thank you so much!!
2. se
(Original post by romina23)
Hey guys, could anyone please explain how to do this question, i end up with +ln2 at the end so I can’t put it into a fraction:

Using the substitution u=e^x +1 show that the integral of e^(2x)/(e^x +1) with limits 1 to 0 is equal to e - 1 - ln((e+1)/2)

(Question 6 from the June 2006 OCR paper in case that’s easier to look at)

Thank you so much!!
can you write down the whole expression you got for an answer please
3. (Original post by yusyus)
se

can you write down the whole expression you got for an answer please
Yeah sure I got

e + 1 - ln(e + 1) - 2 + ln2

Which I simplified to

e - 1 + ln2 - ln(e + 1)

But that means that before the fractions it would be + rather than -

Thank you!!
4. (Original post by romina23)
Yeah sure I got

e + 1 - ln(e + 1) - 2 + ln2

Which I simplified to

e - 1 + ln2 - ln(e + 1)

But that means that before the fractions it would be + rather than -

Thank you!!
that's exactly what I got, now if you put the logs together you get
e-1 + ln( 2/e+1)

if you want e+1/2 think about what power you will have to raise what is currently in the log to get that, if that makes sense
5. (Original post by yusyus)
that's exactly what I got, now if you put the logs together you get
e-1 + ln( 2/e+1)

if you want e+1/2 think about what power you will have to raise what is currently in the log to get that, if that makes sense
Oh could you do e to the power of each term and cancel somethings down?

EDIT: oh wait you wouldn’t be able to because you have e terms anyway, ahhh I’m so confused
6. (Original post by romina23)
Oh could you do e to the power of each term and cancel somethings down?

EDIT: oh wait you wouldn’t be able to because you have e terms anyway, ahhh I’m so confused
I didn't explain it very well, basically.

if you have a fraction x/y and you put the whole thing to the power of -1 (x/y)^-1 you will get y/x.

The same can be applied here; you have e-1 + ln (2/e+1). if you then raise whats inside the log to the power of -1

e-1 + ln(e+1/2)^-1

then use log rules to bring the negative power down
7. (Original post by yusyus)
I didn't explain it very well, basically.

if you have a fraction x/y and you put the whole thing to the power of -1 (x/y)^-1 you will get y/x.

The same can be applied here; you have e-1 + ln (2/e+1). if you then raise whats inside the log to the power of -1

e-1 + ln(e+1/2)^-1

then use log rules to bring the negative power down
Ohhhh i understand now - never would’ve thought of that haha, thank you so so much!!!

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Updated: February 13, 2018
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