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Integration by parts MIDTERM really quick question please help

  1. Feb 2, 2009 #1
    integration by parts!!!! MIDTERM...really quick question....please help!!!!

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

    Evaluate the two following integrals:
    [tex]\int[/tex] xcos5x dx and [tex]\int[/tex] ln(2x+1) dx

    2. Relevant equations



    3. The attempt at a solution

    ok, for the first one, the answer is 1/5 x sin5x + 1/25cos5x+C

    i get this answer up until the 1/5x and the 1/25....i have no idea where there coming from. in my solutions manual they come up when u take the integral of cos5x....its 1/5sin5x....where is this 1.5 coming from?

    secondly the answer to the second one is: 1/2(2x+1)ln(2x+1)-x+C

    again i get this exact answer except for the 1/2. i know its coming from when u take the integral of 1/(2x+1).....but i dont get where the 1/2 comes from in front of the natural log

    help would be greatly appreciated...i have a midterm wednesday!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 2, 2009 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    chain rule

    Hi banfill_89! :smile:

    It's the chain rule (backwards!) …

    differentiate sin5x and you get an extra 5 because of the chain rule … same with ln(2x+1), you get an extra 2 …

    so when you integrate, you have to cancel them out with a 1/5 or 1/2. :wink:
     
  4. Feb 2, 2009 #3
    Re: integration by parts!!!! MIDTERM...really quick question....please help!!!!

    ohhhhhhh. makes alot of sense....thanks alot
     
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