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Simple integration by parts problem

  1. Jan 29, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]
    \int \ln(2x+1)dx
    [/tex]

    2. Relevant equations


    3. The attempt at a solution

    [tex]u = \ln (2x +1)[/tex]

    [tex]du = \frac{2}{2x+1}[/tex]

    [tex]dv = dx[/tex]

    [tex]v = x[/tex]


    [tex]
    xln(2x+1) - \int \frac {2x}{2x+1}dx
    [/tex]


    I'm not sure how to proceed. Do I separate the fraction in the integrand or do long division?
    I think I separate the fraction but I don't remember how. So I guess the question is how do I separate the fraction in the integrand?
     
  2. jcsd
  3. Jan 29, 2010 #2

    ideasrule

    User Avatar
    Homework Helper

    You can rewrite the numerator as 2x+1-1, then separate the fraction into 1-1/(2x+1). A slightly easier way of doing this problem would be to rewrite the original integral as [tex]1/2\int \ln(u)du[/tex], with u=2x+1, then integrating ln(u) by parts.
     
  4. Jan 29, 2010 #3
    Thank you!

    So now I have:

    [tex]
    \int \ln(2x+1)dx = x\ln(2x+1) - \int (1 - \frac{1}{2x+1})dx
    [/tex]

    I'm stuck on the next step. I think I need to remove the 1 somehow from in front of the fraction.
     
  5. Jan 29, 2010 #4

    Mark44

    Staff: Mentor

    Continuing where you left off:
    [tex]\int \ln(2x+1)dx = x\ln(2x+1) - \int (1 - \frac{1}{2x+1})dx[/tex]
    [tex]=x ln(2x + 1) - x + \int \frac{dx}{2x + 1}[/tex]
    The -x term came from integrating -1 with respect to x. The last integral can be done with a simple substitution, u = 2x + 1, du = 2dx.

    Don't forget the constant of integration.
     
  6. Jan 30, 2010 #5
    Perfect! Thank you.
     
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