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Basic Integral

  1. Oct 25, 2006 #1

    danago

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    Hey. Say i was given this indefinite integral to evaluate:

    [​IMG]

    How could i do that? I can do it by first expanding it all, but that takes a very long time and is quite tedious, especially with such a large index as 7. Is there another way i can do that?

    Thanks,
    Dan.
     
  2. jcsd
  3. Oct 25, 2006 #2

    StatusX

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    You could use substitution.
     
  4. Oct 25, 2006 #3

    danago

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    What would i substitute out? I tried using substitution, but couldnt get very far.
     
  5. Oct 25, 2006 #4

    danago

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    [​IMG]

    Where from there?
     
  6. Oct 25, 2006 #5

    HallsofIvy

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    Why squared? that leaves a fraction power. I would just let
    u= 3x+2 then du= 3dx so dx= (1/3)du. Also, x= u/3+ 2/3 so
    x(3x+2)7dx= (u/3+ 2/3)(u7)(1/3du)= (1/9)(u8+ 2u7)dx

    Danago, you have remember to replace dx with du. If u= (3x+2)2, then du= 2(3x+2)dx= (6x+ 4)dx. I think courtrigrad's point was that you can use that "x" in the integeral to help with that. But with that "4dx" still left, I think u= 3x+2 is simpler.
     
  7. Oct 25, 2006 #6
    doesn't x = u/3 - 2/3
     
  8. Oct 25, 2006 #7

    danago

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    Hmmm...im a bit lost. I understand up to "dx= (1/3)du", but where does the "x= u/3+ 2/3" come from? If it is anything more advanced than substitution, i think ill leave it there, because i just made this question up out of curiosity, not because i need to know it for school, and even substitution is more advanced than what we've been doing at school, but i learnt it to make some of the question we do simpler.
     
  9. Oct 25, 2006 #8
    If u = 3x +2, then x = u/3 - 2/3. You have to solve for x.
     
  10. Oct 25, 2006 #9

    danago

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    oh lol. That simple :P
     
  11. Oct 25, 2006 #10

    danago

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    Is this right?
    [​IMG]
     
  12. Oct 25, 2006 #11
    Yes it's correct, but you can tidy it up a little.
     
  13. Oct 25, 2006 #12

    danago

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    ok thanks everyone :)
     
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