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    Name:  image.jpg
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Size:  507.5 KBHey guys,

    This question is from the AQA c3 June 13 paper (see attached image).

    I understand how to do a) i), and hence also tried to integrate (ln(x))^2 by parts in ii) with no success. I set u=lnx and dv/dx=ln x also, but according to the mark scheme I'm completely wrong in doing so.

    Could anybody please explain how I should approach this question?

    Thanks
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    (Original post by Serine)
    Could anybody please explain how I should approach this question?
    Hey!

    The first part asks you to integrate \int \ln x \, \mathrm{d}x by setting u = \ln x and \mathrm{d}v = 1.

    So, it is by no means, a big leap to integrate \int (\ln x)^2 \, \mathrm{d}x by setting u = (\ln x)^2 and \mathrm{d}v = 1.

    Essentially, what the question is demonstrating in the first part is that "set u = ugly logarithm" and "dv = something nice and simple" where you can see that v then becomes x and du becomes something * 1/x. This then cancels in the integral of v du part of your IBP.

    So, when you get to (ii), you should be able to see that setting u = (ln x)^2 still gets you du = something * 1/x and dv becomes x so that integral of v du becomes something simple and cancels. Basically, try and never have dv = a logarithm function. That's plain hell.

    Edit: Whilst the above is what I'd use personally, since it links back to the previous part nicely - I see no reason why your method wouldn't work. Is there a reason you think it doesn't work?
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    (Original post by Serine)
    Name:  image.jpg
Views: 82
Size:  507.5 KBHey guys,

    This question is from the AQA c3 June 13 paper (see attached image).

    I understand how to do a) i), and hence also tried to integrate (ln(x))^2 by parts in ii) with no success. I set u=lnx and dv/dx=ln x also, but according to the mark scheme I'm completely wrong in doing so.

    Could anybody please explain how I should approach this question?

    Thanks
    By parts twice
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    You can set u=\ln(x) and \frac{dv}{dx}=\ln(x) in integration by parts. Then use the first part.

    Edit:

     \frac{du}{dx} = \frac{1}{x}, v=x\ln(x)-x

    [\ln(x)(x\ln(x)-x)] - \int (x\ln(x)-x)(\frac{1}{x}) dx

    (x\ln(x)-x)(\frac{1}{x}) = \ln(x)-1 which you can integrate using the first part.
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    (Original post by Zacken)
    Hey!

    The first part asks you to integrate \int \ln x \, \mathrm{d}x by setting u = \ln x and \mathrm{d}v = 1.

    So, it is by no means, a big leap to integrate \int (\ln x)^2 \, \mathrm{d}x by setting u = (\ln x)^2 and \mathrm{d}v = 1.

    Essentially, what the question is demonstrating in the first part is that "set u = ugly logarithm" and "dv = something nice and simple" where you can see that v then becomes x and du becomes something * 1/x. This then cancels in the integral of v du part of your IBP.

    So, when you get to (ii), you should be able to see that setting u = (ln x)^2 still gets you du = something * 1/x and dv becomes x so that integral of v du becomes something simple and cancels. Basically, try and never have dv = a logarithm function. That's plain hell.
    Thank you yet again for the help!!
    Is it correct to say when IBP that dv shouldn't be the same as u? (I usually use LIATE to find dv and u).
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    (Original post by Serine)
    Thank you yet again for the help!!
    Is it correct to say when IBP that dv shouldn't be the same as u? (I usually use LIATE to find dv and u).
    Erm, not particularly. You could integrate \int x^2 \, \mathrm{d}x with u = \mathr{d}v = x.

    By the way - your method works out just fine! It's perfectly valid! Setting u = \mathrm{d}v = \ln x works out as well. Why do you think it doesn't? :-)
 
 
 
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