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Integration by parts VS. Formula in text

  1. Oct 28, 2007 #1
    I am either making the same mistake repeatedly, or I can't factor!

    I did this [tex]\int\ln(2x-3)dx[/tex] first by parts and then using the formula [tex]\int\ln u du=u\ln u-u+C[/tex]

    By parts I got:
    [tex]u=\ln(2x-3)[/tex] dv=dx

    so [tex]=x\ln(2x-3)-\int\frac{2x}{2x-3}dx[/tex] and by long division:

    [tex]=x\ln(2x-3)\int [1+\frac{3}{2x+3}]dx[/tex]


    but I can't get that to match the formula result of


    Is the by parts correct?

  2. jcsd
  3. Oct 28, 2007 #2
    I am getting closer, but I must have screwed up a sign or something in the Int by parts.....but I cant see where .......
  4. Oct 28, 2007 #3
    that's what i got, let me take the derivative and to check.
    Last edited: Oct 28, 2007
  5. Oct 28, 2007 #4
    Okay, so I have fixed a sign error and I get from parts:

    [tex]x\ln(2x-3)-x-\frac{3}{2}\ln(2x-3)[/tex] and then multiplying thru

    by 2 I get [tex]2x\ln(2x-3)-2x-3\ln(2x-3)[/tex] which is extremely close...but I can't figure out how to factor it properly....

    Almost there :)

    I just can't figure out to do with that -2x in the "middle" its a term not a factor......
  6. Oct 28, 2007 #5
    For one,

    [tex]x\ln(2x-3)\int [1+\frac{3}{2x+3}]dx[/tex]

    should have a minus sign within the integral.

    In the following step, you've multiplied the - outside the integral(that's supposed to be there :wink:) with that +.
  7. Oct 28, 2007 #6
    it checks out after taking the derivative.
  8. Oct 28, 2007 #7
    I don't follow. If I do that integral off to the side I get x+(3/2)ln(2x-3)

    and then i distibuted the - sign.....

    EDIT: I see, I just forgot to put it in the 1st post...scroll down to post#4
    Last edited: Oct 28, 2007
  9. Oct 28, 2007 #8
    What does? from which post#. I can't get it to match the formula.

  10. Oct 28, 2007 #9
    first post, except that the last ln should be ln(2x-3)

    i hate the tables, i prefer hand-method :p takes too long and comes with errors cuz i suck, but is all good.
  11. Oct 28, 2007 #10
    My goal though is to get what I got in post #4 to look like the formula from the text book which looks like this [tex](2x-3)\ln(2x-3)-(2x+3)+C[/tex]

    but I have [tex]2x\ln(2x-3)-2x-3\ln(2x-3)+C[/tex]

    which is close...how does that factor?

    If I factor that don't I just get [tex] (2x-3)\ln(2x-3)-2x+C[/tex] ?
    Last edited: Oct 28, 2007
  12. Oct 28, 2007 #11
    why can't I factor this:rofl:??!
  13. Oct 28, 2007 #12
    lol that's really hard ... quit obsessing!
    Last edited: Oct 28, 2007
  14. Oct 28, 2007 #13
    But it is part of the assignment!
  15. Oct 28, 2007 #14
    to put it in table form? eek!

    i got:

    [tex]\ln{(2x-3)}[2x-3]-x[/tex] ...

    but idk what to do after that. and how did you get -2x?
  16. Oct 28, 2007 #15
    Maybe it's because what you're trying to find is an indefinite integral, and the answers to different methods employed usually differ by a constant.
  17. Oct 29, 2007 #16
    This might help:


    In which [tex]C_1=C-\frac{3}{2}[/tex] another constant is.
  18. Oct 29, 2007 #17


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    Homework Helper

    salad, I can't quote your post, browser issues. But your "book formula" integral is in error.


    should read


    (I corrected your sign error)

    You cannot in general just use a formula for an integral of f(u) with a direct substitution of u=g(x) into the result. This is because what you are actually doing with a proper substitution is

    [tex]\int{f(g(x))}dx = \int{\frac{dx}{du}[f(u)]}du = \int{\frac{f(u)}{u'}}du[/tex]

    In this case, u'(x) = 2, so you should divide the final expression by two.

    I did not check your integration by parts, but the above correct answer for the formula method tallies with it, except for a constant term :

    [tex]x\ln{(2x-3)} - x - \frac{3}{2}\ln{(2x-3)} = \frac{1}{2}(2x)\ln{(2x-3)} - \frac{1}{2}(2x) - \frac{1}{2}(3\ln{(2x-3)}) = \frac{1}{2}[(2x-3)\ln{(2x-3) - 2x}] = \frac{1}{2}[(2x-3)\ln{(2x-3) - (2x - 3)}] + C[/tex]

    which all works out.
    Last edited: Oct 29, 2007
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