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# Integration- Core 3 watch

1. How would I integrate 4 / x(4lnx + 3)^2, with the limits of e and 1. I know that dy/dx of lnx= 1/x but don't know how to integrate it?

Also, when using modulus such as l 4x -3 l < l 2x + 1 l. I can't seem to get the right answer. Is the method squaring both sides when there are inequalities of modulus, right?
2. Do you mean differentiation? Integration isn't in C3. What question/paper is this?
For the 2nd question, I wouldn't square it cos I dunno if you can but also it loses some of the answers if gets any of them.
l4x-3l = 4x-3 AND -4x+3
this is the same for the 2nd side: l2x+1l = 2x+1 AND -2x-1
therefore the are 4 combinations of the overall equation... long, i know
GOOD LUCK TOMORROW!
3. Could you link towards the question itself? You might have overlooked a small comment or whatever. I personally always prefer to get the question at hand, since I've seen people misread stuff too many times to puzzle my head over something that might have been a lot simpler.
4. (Original post by phen)
Could you link towards the question itself? You might have overlooked a small comment or whatever. I personally always prefer to get the question at hand, since I've seen people misread stuff too many times to puzzle my head over something that might have been a lot simpler.

Questions 2 and 8iii. For the first part, I think I know understand that you make the different possible combinations, but when does squaring both sides work and when does it not?
5. (Original post by definite_maybe)
How would I integrate 4 / x(4lnx + 3)^2, with the limits of e and 1. I know that dy/dx of lnx= 1/x but don't know how to integrate it?
If the question's , try a substitution of .
6. Have a look at your answer to part 8(i) for a massive hint.

(Original post by james.h)
If that's the question, try a substitution of .
Integration by substitution doesn't come up in the OCR syllabus until C4; however there's enough information needed to do the integration in the first part of the question.
7. (Original post by nuodai)
Have a look at your answer to part 8(i) for a massive hint.

Integration by substitution doesn't come up in the OCR syllabus until C4; however there's enough information needed to do the integration in the first part of the question.
With the 4lnx-3/4lnx+3, to get the value integrated do you divide by 6 or multiply by 6? Both times I get the wrong answer?

Also for the modulus, do you only use squared when it is =, and not for inequalities?
8. Yo,

im new to this and im not sure how to type it all out...but here goes.

V = pi ∫(e to 1) y^2 dx

= pi ∫(e to 1) 4/x(4lnx+3)^2 dx

= pi/6 ∫ (e to 1) 24/x(4lnx+3)^2 dx

= pi/6 [4lnx-3/4lnx+3] (e to 1)

= pi/6 ((1/7 - (-1))

= pi/6 x (1/7 + 1)

= pi/6 x 8/7 (Cancel This Down)

= 4pi/21
9. (Original post by Bhav_Singh)
Yo,

im new to this and im not sure how to type it all out...but here goes.

V = pi ∫(e to 1) y^2 dx

= pi ∫(e to 1) 4/x(4lnx+3)^2 dx

= pi/6 ∫ (e to 1) 24/x(4lnx+3)^2 dx

= pi/6 [4lnx-3/4lnx+3] (e to 1)

= pi/6 ((1/7 - (-1))

= pi/6 x (1/7 + 1)

= pi/6 x 8/7 (Cancel This Down)

= 4pi/21
Thank you. Any help with the modulus?
10. If you draw them both out, it helps.

But ill just show you how to get the answer but then just try drawing them out and getting the answer.

4x-3 = 2x+1

2x = 4

x = 2

2x+1 = 3-4x

6x = 2

x = 1/3

Therefore, 1/3 < x < 2
11. (Original post by Bhav_Singh)
If you draw them both out, it helps.

But ill just show you how to get the answer but then just try drawing them out and getting the answer.

4x-3 = 2x+1

2x = 4

x = 2

2x+1 = 3-4x

6x = 2

x = 1/3

Therefore, 1/3 < x < 2
I udnerstand how to get them by graph method, but what I don't understand is in which situations you can use squaring to solve
12. Lol, i dont know. I dont do OCR anyway, I do MEI - so its a bit different.

gl with your C3 exam tomorrow. X

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