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Homework Help: Def Integral of ln(x+2)dx

  1. Oct 26, 2007 #1
    So I am using int by parts to evaluate [tex]\int_{-1}^1\ln(x+2)dx[/tex]

    I am doing something incorrectly with the indefinite int part of this problem, but I can'y find my error, can you?

    [tex]\int udv=uv-\int vdu[/tex]
    [tex]u=ln(x+2)[/tex]
    [tex]\Rightarrow du=\frac{dx}{x+2}[/tex]
    [tex]dv=dx[/tex]
    [tex]\Rightarrow v=x[/tex]

    So I get:
    [tex]x\ln(x+2)-\int\frac{x}{x+2}dx[/tex]

    Working on only the last part ^^^by long division I get:
    [tex]\int\frac{x}{x+2}dx=\int(1-\frac{2}{x+2})dx[/tex]

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

    and recombining with the top:

    [tex]x\ln(x+2)-x+2\ln(x+2)[/tex]

    But when I evaluate this on my calculator, I get the incorrect answer. Any ideas?

    Casey
     
    Last edited: Oct 26, 2007
  2. jcsd
  3. Oct 26, 2007 #2
    Why use integration by parts? Why not just use substiution?

    u=x+2
    du/dx=1
     
  4. Oct 26, 2007 #3
    Because that is not the point of the exercise.:biggrin: A u-sub will do no good if we have not yet learned the integral of lnu

    Casey

    But, nonetheless, my integral is correct. I am doing something incorrectly with my calculator.

    edit: I figured it out. damn. what a waste of typing
     
    Last edited: Oct 26, 2007
  5. Oct 26, 2007 #4
  6. Oct 26, 2007 #5
    The answer seems fine to me...

    Use the distributive law to 'pull' the ln(x+2) out.
     
  7. Oct 26, 2007 #6
    yeah i also got the same thing,
     
  8. Oct 26, 2007 #7
    wow....yeah. The things I forget sometimes:redface:

    Thanks guys.

    BTW, it is common too use, as Wolfram does, LOG instead of LN....I thought that Log implied base 10 always?
     
  9. Oct 26, 2007 #8
    To each his own. It's best to use what is being used in your course/textbook, at least for now.
     
  10. Oct 26, 2007 #9
    Well, this depends pretty much on the author. Here at my country when we write only Log we assume it to be with base 10, and Ln to be the natural logarithm with base e. However i have seen some proffesors at some universities usually at US universities often use Log to mean Ln.
     
  11. Oct 26, 2007 #10
    'Pure' mathematicians will generally have log denote the logarithm with base e. This is natural as all power functions (a^x) are really just scaled exponentials and we have a similar result for logs as a consequence.
     
  12. Aug 1, 2008 #11
    I 've been trying to work out an integral similar to the one above but i cant seem to get it correct
    [tex]\int \ln(x+x^{2}) dx[/tex]

    [tex]u = \ln(x+x^{2})[/tex]
    [tex]du = dx/(x+x^{2})[/tex]

    [tex]dv = dx[/tex] : [tex]v = x[/tex]

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

    To figure out the last part of the integral for some reason has me stuck :(
     
    Last edited: Aug 1, 2008
  13. Aug 1, 2008 #12

    Defennder

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    Your du is incorrect.
     
  14. Aug 1, 2008 #13
    ok. but the differential of [tex] \ln(x) = 1/x [/tex]
     
  15. Aug 1, 2008 #14

    Defennder

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    But the expression inside the ln is x^2+x and not x. So you have to apply the general rule for differentiating ln(f(x)).
     
  16. Aug 1, 2008 #15
    mj12, you forgot the most important rule of differential calculus ;)
     
  17. Aug 1, 2008 #16
    By General rule you mean :ln f(x) = f`(x) / f(x)

    which would give me 1 + 2x / x + x^2
     
  18. Aug 1, 2008 #17

    Defennder

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  19. Aug 1, 2008 #18
    Ahhh, your notation had me confused for a second. If you mean (1+2x)/(x+x^2), you are correct. Now multiply that by x (to get vdu) and see if you can integrate that (it's a bit tricky).
     
  20. Aug 1, 2008 #19

    D H

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    Try factoring [itex]x+x^2[/tex] and then use the identity [itex]\log(ab) = \log a + \log b[/itex].
     
  21. Aug 1, 2008 #20

    Defennder

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    That surprisingly, is the harder way to do it, mostly because it's easier to remember the anti-derivative of f'(x)/f(x) than that of ln(x).
     
  22. Aug 1, 2008 #21
    Tricky indeed because i still havent got it right.......I think i need a break from Calc, been at it for a couple of hours now, didnt even notice the thread had a page 2 :|
     
  23. Aug 1, 2008 #22
    Ok well after multiplying by x you get (x+2x^2)/(x+x^2), factoring out an x and canceling gives (2x+1)/(x+1). Now we can manipulate this expression a bit. What happens if you add and subtract 1 from the numerator? In one sense this doesn't change the expression since it's just adding 0 but notice that you get a term in the numerator that is a multiple of the denominator.
     
    Last edited: Aug 1, 2008
  24. Aug 1, 2008 #23
    ok and that would give me 1/(x+1) - 2 ?
    which would become ln(x+1) -2x
     
  25. Aug 1, 2008 #24
    Hmm (2x+2-1)/(x+1) = 2 - 1/(x+1), so you should get 2x - ln(x+1). Don't forget the constant and differentiate to check of course.

    Wait sorry, you must've accounted for the -1 factor already...
     
  26. Aug 1, 2008 #25
    Oops, thats what i mean :P so

    [tex]\int \ln(x+x^{2}) dx[/tex]

    [tex]u = \ln(x+x^{2})[/tex]
    [tex]du = (1 +2x)/(x+x^{2})[/tex]

    [tex]dv = dx[/tex] : [tex]v = x[/tex]

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

    = [tex]x \ln (x+x^{2}) - 2x + \ln (x + 1) + C [/tex]

    Thanks for your help guys really appreciate it. Tomorrow I tackle

    [tex]\int (sin^{-1} x)^{2} dx[/tex]
     
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