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# P3 Integration Dude's Hard One! Who Can Crack It?! watch

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1. question:

integral of (lnx)^2dx

with limits of e and 1
2. (Original post by mabs)
question:

integral of (lnx)^2dx

with limits of e and 1
is the answer 2 by any chance?!
3. e-2
4. (e - 2) ?
5. what method is everyone using - i don't know how to start
6. (Original post by mabs)
question:

integral of (lnx)^2dx

with limits of e and 1
2-ln1
7. (Original post by Silly Sally)
what method is everyone using - i don't know how to start

lol, intergrate by parts..... 2.lnx
8. (Original post by Silly Sally)
what method is everyone using - i don't know how to start
by parts
u=(lnx)^2
dv/dx = 1
...
9. u g2 use parts
10. (Original post by Bhaal85)
2-ln1
ln1 is 0, therefore answer is 2....which is wot i got, but its not right, lol
11. (Original post by Silly Sally)
what method is everyone using - i don't know how to start
lnx^2 can be done by parts, just do integration by parts with 1*lnx^2
12. IT STILL DOESN'T WORK!

U GET 2/X lnx AND THAT WON'T REDUCE
13. (Original post by mabs)
question:

integral of (lnx)^2dx

with limits of e and 1
If you fell up to it you can integrate by parts

u = lnx
u' = 1/x
v = xlnx + x
v' = lnx

integral of question = [ x(lnx)^2 - xlnx ] - [ integral of (lnx -1) ]

=> x(lnx)^2) - xlnx - xlnx +x +x

then simplify and put the limits in and you get e-2. Maybe someone with html can type it out so it looks similar. Im sure theres an easier method but this one works!
14. check P3 ex 4d q.19
15. I got the right answer so here's the method:
Integrate by parts with v = lnx, dv/dx = 1/x, du/dx = lnx, u = x(lnx - 1)
So int (lnx)^2 = xlnx(lnx - 1) - int 1/x*(x(lnx - 1)) = x[(lnx)^2 - lnx] - x(lnx -1) + x = x[ln(x)^2 - 2 lnx + 2], evaluate between 1 and e to give (e - 2)
16. INT lnx^2 dx = INT 1*lnx^2

u = lnx^2 therefore u' = 2/x
v' = 1 therefore v = x

xlnx^2 - INT x*2/x which gave me:

[xlnx^2 - 2x] with limits e and 1

[2e-2e] - [ln1-2]

= 2-ln1
= 2
17. George Explain!!!!!
18. BHAAL, (lnx)^2 means lnx all to the power 2 and u don't get wot u get when u use the chain rule.
19. (Original post by mabs)
BHAAL, (lnx)^2 means lnx all to the power 2 and u don't get wot u get when u use the chain rule.
It does work if you use the chain rule - thats the only bit i can do!!!
20. (Original post by Bezza)
I got the right answer so here's the method:
Integrate by parts with v = lnx, dv/dx = 1/x, du/dx = lnx, u = x(lnx - 1)
So int (lnx)^2 = xlnx(lnx - 1) - int 1/x*(x(lnx - 1)) = x[(lnx)^2 - lnx] - x(lnx -1) + x = x[ln(x)^2 - 2 lnx + 2], evaluate between 1 and e to give (e - 2)

i didnt know u could intergrate 'ln' functions?! how does du/dx + lnx become u = x(lnx - 1)???

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