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    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!!
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    (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
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    (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!!
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    (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
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    (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:confused:
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    (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: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
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    (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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