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Intergrating (ln(x))^2

image.jpgHey 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 :smile:
Reply 1
Original post by Serine
Could anybody please explain how I should approach this question?


Hey!

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

So, it is by no means, a big leap to integrate (lnx)2dx\int (\ln x)^2 \, \mathrm{d}x by setting u=(lnx)2u = (\ln x)^2 and dv=1\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?
(edited 7 years ago)
Reply 2
Original post by Serine
image.jpgHey 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 :smile:


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

Edit:

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

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

(xln(x)x)(1x)=ln(x)1(x\ln(x)-x)(\frac{1}{x}) = \ln(x)-1 which you can integrate using the first part.
(edited 7 years ago)
Reply 4
Original post by Zacken
Hey!

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

So, it is by no means, a big leap to integrate (lnx)2dx\int (\ln x)^2 \, \mathrm{d}x by setting u=(lnx)2u = (\ln x)^2 and dv=1\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!! :smile:
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). :smile:
Reply 5
Original post by Serine
Thank you yet again for the help!! :smile:
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). :smile:


Erm, not particularly. You could integrate x2dx\int x^2 \, \mathrm{d}x with
Unparseable latex formula:

u = \mathr{d}v = x

.

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

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