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Integration by Parts using Ln(x)

  1. Jun 8, 2013 #1
    1. The problem statement, all variables and given/known data

    [tex]\int_4^5 \frac{dx}{(3x)(ln(x))(ln^3(ln(x)))} \,[/tex]

    2. Relevant equations

    I'll probably need to use u-substitution as well as integration by parts for this problem.


    3. The attempt at a solution
    [tex]\int_4^5 \frac{dx}{(3x)(ln(x))(ln^3(ln(x)))} \,[/tex]

    1. Factor out 1/3 for convenience.

    [tex]\frac{1}{3}\, \int_4^5 \frac{dx}{(x)(ln(x))(ln^3(ln(x)))} \,[/tex]


    2. Let u = ln(x), du = 1/x * dx, thus dx = du * x


    [tex]\frac{1}{3}\, \int_4^5 \frac{(x)(du)}{(x)(u)(ln^3(u))} \, = \frac{1}{3}\, \int_4^5 \frac{du}{(u)(ln^3(u))} \,[/tex]

    From integration by parts we know that ∫g*dv = g*v - ∫dg*v

    I'm stuck here with regards to what to assign as g and what to assign as dv.

    I could make:
    g = 1/u
    dg = -1/u^2

    dv = 1/ln^3(u) * du
    v = ????

    I'm not sure how to integrate dv to get to v to make the integration by parts work.

    Any help would be appreciated.
     
  2. jcsd
  3. Jun 8, 2013 #2
    Make a second substitution
     
  4. Jun 8, 2013 #3

    lurflurf

    User Avatar
    Homework Helper

    Integration by parts is not helpful here

    You wither want two use the substitution

    u=(log log x)^3

    or equivalently another substitution after yours

    u=log x
    w=(log u)^3
     
  5. Jun 8, 2013 #4
    That second substitution complicates the problem, a simple w=lnu is enough
     
  6. Jun 8, 2013 #5
    Thanks for your help, guys. I think I understand what you're saying.

    Let g = ln(u), dg = 1/u * du, thus du = dg* u

    Substituting back in I get:


    [tex]\frac{1}{3}\, \int_4^5 \frac{dg*u}{(u)(g^3)} \, = \frac{1}{3}\, \int_4^5 \frac{dg}{g^3} \, = -\frac{1}{2g^2}\,\frac{1}{3}\, [/tex]

    and then I would substitute back to reobtain x and apply the limit from 4 to 5.

    Thank you, guys. Much appreciated.
     
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