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    http://qualifications.pearson.com/co...e_20150612.pdf

    how would I do part 13C, im confused
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    Which question?
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    (Original post by imran_)
    http://qualifications.pearson.com/co...e_20150612.pdf

    how would I do part C, im confused
    Which question?

    And please post your working, thoughts and reason why you're stuck.
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    (Original post by NotNotBatman)
    Which question?
    13c sorry
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    (Original post by notnek)
    Which question?

    And please post your working, thoughts and reason why you're stuck.
    13c, its asking to work it out by parts but I have no clue where to start. What I did first was bring the 2 down to get 2lnx and solved from there but its wrong
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    Multiply the natural log by 1.
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    (Original post by imran_)
    13c, its asking to work it out by parts but I have no clue where to start. What I did first was bring the 2 down to get 2lnx and solved from there but its wrong
    Try integrating by calling it 1*(Lnx)^2 with 1 being your v'.
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    (Original post by imran_)
    http://qualifications.pearson.com/co...e_20150612.pdf

    how would I do part 13C, im confused
    for intergration by parts use the formula. integral of v du/dx= uv + integral of v du/dx and because it's a ln need to make dv/dx 1 and u (lnx)^2 hope that helps
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    So, when you have a natural log function, ln[f(x)], you can integrate it by parts using 1 x ln[f(x)] and choosing your u and dv.
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    (Original post by NotNotBatman)
    Multiply the natural log by 1.
    sorry im confused, i made u=lnx and solved from there and got the correct answer but how would you do your method?
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    (Original post by NotNotBatman)
    So, when you have a natural log function, ln[f(x)], you can integrate it by parts using 1 x ln[f(x)] and choosing your u and dv.
    Are you Batman?
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    (Original post by NotNotBatman)
    So, when you have a natural log function, ln[f(x)], you can integrate it by parts using 1 x ln[f(x)] and choosing your u and dv.
    okay ill give it a go!
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    (Original post by imran_)
    sorry im confused, i made u=lnx and solved from there and got the correct answer but how would you do your method?
    Latex isn't working for me, so it's a bit hard to explain, I'd basically do it the way examsolutions does in this video , except making u =(lnx)^2 .

    (Original post by eden3)
    Are you Batman?
    Spoiler:
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    Well, I'm not not batman.
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    (Original post by NotNotBatman)
    So, when you have a natural log function, ln[f(x)], you can integrate it by parts using 1 x ln[f(x)] and choosing your u and dv.
    Well I don't think you can extend the result to \ln(f(x)) since that will only work if you can integrate \dfrac{xf'(x)}{f(x)}
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    (Original post by NotNotBatman)
    Latex isn't working for me, so it's a bit hard to explain, I'd basically do it the way examsolutions does in this video , except making u =(lnx)^2 .
    Spoiler:
    Show
    Well, I'm not not batman.
    okay yeah I just worked it out and gave me the right answer, thanks!
    However for part D of the question y=(lnx)^2 if I square it I get (lnx)^4 no?
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    (Original post by eden3)
    Are you Batman?
    Back to chat please
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    (Original post by imran_)
    okay yeah I just worked it out and gave me the right answer, thanks!
    However for part D of the question y=(lnx)^2 if I square it I get (lnx)^4 no?
    No, y= 2-ln(x), part (c) is just to help you for part (d).
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    (Original post by NotNotBatman)
    No, y= 2-ln(x), part (c) is just to help you for part (d).
    i dont get it. how did you manage to get that
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    (Original post by imran_)
    i dont get it. how did you manage to get that
    The question says "figure 5 shows a sketch of part of the curve with equation y = 2 - ln x "
 
 
 
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