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Integration by parts MIDTERM really quick question please help

  1. Feb 3, 2009 #1
    integration by parts!!!! MIDTERM...really quick question....please help!!!!

    1. The problem statement, all variables and given/known data

    [tex]\int[/tex] arctan4t dt


    2. Relevant equations



    3. The attempt at a solution

    ive been attempting this all day. the answer is t arctan4t-1/8 ln(1+16t^2)+C

    i get this asnwer up until the 1/8.......

    a few steps befor emy answer i have t arctan4t - [tex]\int[/tex] 4t/ 1+16t^2

    i dont kno whow to go from here to the answer....someone please help!!!
     
  2. jcsd
  3. Feb 3, 2009 #2

    Mark44

    Staff: Mentor

    Re: integration by parts!!!! MIDTERM...really quick question....please help!!!!

    In your last integral, there's a fairly obvious substitution.

    BTW, you need more parentheses: the integrand can be written more understandably as 4t/(1 + 16t^2)
     
  4. Feb 3, 2009 #3
    Re: integration by parts!!!! MIDTERM...really quick question....please help!!!!

    what would be the substituition? in the book after thsi they factor out a 1/8......im lost
     
  5. Feb 3, 2009 #4

    Dick

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    Science Advisor
    Homework Helper

    Re: integration by parts!!!! MIDTERM...really quick question....please help!!!!

    Substitute u=1+16t^2.
     
  6. Feb 3, 2009 #5
    Re: integration by parts!!!! MIDTERM...really quick question....please help!!!!

    ??? so am i integrating by parts again?
     
  7. Feb 3, 2009 #6

    Dick

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    Science Advisor
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    Re: integration by parts!!!! MIDTERM...really quick question....please help!!!!

    No, it's a u substitution.
     
  8. Feb 3, 2009 #7
    Re: integration by parts!!!! MIDTERM...really quick question....please help!!!!

    i finnaly got it. this textbook sucks at explaining this. thanks guys. sorry dick i was confused at first cause u said let u= 16x^2+1 and i already used u. i get what you were saying now and it makes alot of sense. thanks!
     
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