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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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