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Math Help - integration of natural log

  1. #1
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    integration of natural log

     \int ln(3x - 7)dx
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  2. #2
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    Quote Originally Posted by wik_chick88 View Post
     \int ln(3x - 7)dx
    Set  t= 3x-7 \implies \text{dt}= 3\text{dx}

     \int \ln(3x - 7)\text{dx}  = \frac{1}{3}\int \ln(t)\text{dt}

    u= \ln(t) \implies du= \frac{1}{t}\text{dt}
    \text{dv}= \text{dt} \implies v= t

    \frac{1}{3}\left\{ t \ln(t)- \int t \frac{1}{t} \text{dt} \right\}
    Last edited by Anonymous1; May 10th 2010 at 07:14 AM.
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  3. #3
    Super Member General's Avatar
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    Substitute t=3x-7, to get :
    \frac{1}{3}\int ln(t) \, dt
    Now, use integration by parts ..
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  4. #4
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    the final answer i get is (x - \frac{7}{3})ln(3x - 7) - \frac{ln(3x - 7)}{3} + C
    but the answer given is (x - \frac{7}{3})ln(3x - 7) - x + C

    where did i go wrong??
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  5. #5
    Super Member General's Avatar
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    How can we tell your mistakes if we did not see your solution .. ?!!
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  6. #6
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    Quote Originally Posted by wik_chick88 View Post
    where did i go wrong??
    It would be easier if you post your working so that your mistake can be pointed out, it wasn't helpful with just the final answer posted.
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  7. #7
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by wik_chick88 View Post
    (x - \frac{7}{3})ln(3x - 7) - \frac{1}{3}\color{red}{(3x - 7)} +C

    \color{red}{\text{Here is where you messed up:}}

    \color{red}{\int t\frac{1}{t} = t = (3x-7)}
    aNon1
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  8. #8
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    ok.
    y = \int ln(3x - 7) dx
    let  t = 3x - 7
    \frac{dt}{dx} = 3,
    so.. dx = \frac{1}{3} dt

    so...
    y = \frac{1}{3} \int ln t dt
    let  u = ln t, so u' = \frac {1}{t}
    let  v' = 1 , so v = t

    so...
    y = \frac{1}{3} (t ln t - \int \frac{1}{t} dt)
    = \frac{1}{3} (t ln t - ln t) + C
    = \frac{t ln t}{3} - \frac{ln t}{3} + C
    = \frac{(3x - 7) ln (3x - 7)}{3} - \frac{ln(3x - 7)}{3} + C
    = (x - \frac{7}{3})ln(3x - 7) - \frac{ln (3x - 7)}{3} + C
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  9. #9
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    thankyou thankyou thankyou! silly me!
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