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Evaluate the integral, integration by parts

  1. Aug 19, 2008 #1
    1. The problem statement, all variables and given/known data
    ∫ln (2x+1) dx


    2. Relevant equations
    ∫u dv=uv- ∫ v du
    integration by parts


    3. The attempt at a solution
    u= ln (2x+1) dv=dx
    du=???? v=x

    ok did i choose the right u and how do i derive it, do i have to use the chain rule



    π²³ ∞ 0° ~ µ ∑ Ω √ ∫ ≤ ≥ ± # … θ φ ψ ω Ω α β γ δ ∂ ∆ ε λ Λ Γ ô
     
  2. jcsd
  3. Aug 19, 2008 #2

    Dick

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    Yes. Use the chain rule.
     
  4. Aug 19, 2008 #3

    nicksauce

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    You did choose the 'right' u, although a substitution like g=2x+1 might make things easier, though it's not necessary. To find du, just apply the chain rule, and use the fact that d/dx(ln(x)) = 1/x.
     
  5. Aug 19, 2008 #4
    ok, i'll try it out
     
  6. Aug 19, 2008 #5
    so i used the chain rule and i did this
    ln (2x+1)= 1/x * (2x+1) * 2= du

    did i do it correctly and du is 4x+4/x
     
  7. Aug 19, 2008 #6
    If [tex]u = ln(2x+1)[/tex] then [tex]du = \frac{2 dx}{2x+1}[/tex]
     
    Last edited: Aug 19, 2008
  8. Aug 19, 2008 #7
    so i did it wrong
     
  9. Aug 19, 2008 #8
    i don't get how you got 2/2x+1
     
  10. Aug 19, 2008 #9
    Looks like it :-/

    It might make sense to do the derivative like this:

    First: g(x)=2x+1 and f(x)=ln(g(x)) so then the chain rule is:

    [tex]\frac{d}{dx}f(x) = g'(x)*f'(g(x)) [/tex]

    g'(x) = 2 and f'(g(x)) = 1/g(x) = 1/(2x+1)

    From there is should be pretty easy to follow.
     
  11. Aug 19, 2008 #10
    oh ok, i didn't know i could've used g for 2x+1 that's what the other guy was telling me, i'm really rusty on my calculus, thanks
     
  12. Aug 19, 2008 #11
    this is what i got so far
    u= ln (2x+1) dv=dx
    du= 2/2x+1 v=x

    ln (2x+1) (x) - ∫x (2/2x+1

    now i'm integrating ∫x (2/2x+1 dx
    u=x dv=2/2x+1
    du=dx v=?????
    I'm stuck here how do i antiderive 2/2x+1
     
    Last edited: Aug 19, 2008
  13. Aug 19, 2008 #12
    is it 1/2 (2x+1)
     
  14. Aug 19, 2008 #13
    ok i've decided to use u=2/2x+1 dv=x
    du=2x-4/(2x+1)^2 v=dx
     
  15. Aug 19, 2008 #14
    i forgot about factoring do i have to factor du now
     
  16. Aug 19, 2008 #15

    Dick

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    If you want to integrate (2x)/(2x+1) don't use parts. Use the substitution u=2x+1 again. So 2x=u-1. Can you put it all together?
     
  17. Aug 19, 2008 #16
    ok i'll give it a shot
     
  18. Aug 19, 2008 #17
    ok what do i do after
     
  19. Aug 19, 2008 #18

    Dick

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    After what? What you done so far? I'm just suggesting you do it as a u-substitution. You've done that before, right? What's the integral in terms of u?
     
  20. Aug 19, 2008 #19
    du is the term
     
  21. Aug 19, 2008 #20

    Dick

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    ??? You have 2x/(2x+1)dx. You want to replace all of the parts with their equivalents in terms of u. What's the new u integral?
     
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