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Integration, u substitution with limits

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate ∫ 3x /(3x+1)^2.dx , with limits 1 & 0
    using the sustitution u = 3x+1




    2. Relevant equations




    3. The attempt at a solution

    u= 3x+1
    du/dx = 3
    dx = du/3

    Therefore,

    ∫ 3x*(u)^-2 * du/3

    = ∫ x* (u)^-2

    Since u = 3x +1
    Therefore, x = (u-1)/3

    Hence,

    ∫ (u-1)* 1/3*(u^2)

    Now, plugging in limits of 1&0 into u

    u = 3(1) +1 =4
    u = 3(0) + 1 = 1

    Therefore, limits of 4 & 1.

    Hence,
    1/3 ∫ (u^-1)-(u^-2)

    =1/3 [-u + 2u^-1] with limits 4& 1.

    = 1/3 (- 9/2) = -3/2.

    However, there is no answer mathing this solution.
    But i dont know where i went wrong??
    please help.
    thankyou.
     
  2. jcsd
  3. Feb 21, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Your integral of [tex]\int_1^4{(\frac{1}{u}-\frac{1}{u^2})du}[/tex] is wrong. Look after the basic integral formulae.


    ehild
     
  4. Feb 21, 2012 #3
    Should it be...

    ln u + u^-1
     
  5. Feb 21, 2012 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, it is perfect now!

    ehild
     
  6. Feb 21, 2012 #5
    yay!!

    Thanks so much for the help.
    Greatly appreciated
     
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