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c4 integration help

http://qualifications.pearson.com/content/dam/pdf/International%20Advanced%20Level/Mathematics/2013/Exam%20materials/WMA02_01_que_20150612.pdf

how would I do part 13C, im confused
(edited 7 years ago)
Which question?
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Notnek
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Original post by imran_

Which question?

And please post your working, thoughts and reason why you're stuck.
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imran_
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Original post by NotNotBatman
Which question?


13c sorry
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imran_
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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
Multiply the natural log by 1.
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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dgj14944
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Original post by imran_


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
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!
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

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))\ln(f(x)) since that will only work if you can integrate xf′(x)f(x)\dfrac{xf'(x)}{f(x)}
(edited 7 years ago)
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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



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?
Original post by eden3
Are you Batman?


Back to chat please:biggrin:
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
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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