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Homework Help: How do you integrate this?

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex] \int (3x-5)e^{4ln(x)}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    Any ideas?
  2. jcsd
  3. Sep 24, 2009 #2
    rewrite [tex]e^{4ln(x)}[/tex] as [tex]e^{(ln x)4}[/tex] and remember your rules for exponents!
  4. Sep 25, 2009 #3
    does e^(ln(x)) becomes x?
  5. Sep 25, 2009 #4
  6. Sep 25, 2009 #5
    http://img85.imageshack.us/img85/8667/questionl.th.jpg [Broken]

    If the original question is above and I solved the integral above which results in:

    [tex]x^3 - \frac{5x^2}{2} + \frac{C}{e^{4ln(x)}[/tex]

    then I need to divide this by [tex] e^{4ln(x)} [/tex] right??
    I am also given an initial state of y(1) = 5

    the final result that I get is :


    why is this wrong? which step am I not doing it right?

    however web assign marks it as wrong.. am I doing something wrong?
    Last edited by a moderator: May 4, 2017
  7. Sep 25, 2009 #6

    What's [tex]\int_0^x\frac{4}{t}dt[/tex]
  8. Sep 25, 2009 #7
    Easy mate
    integration of (3x-5) = 3x^2-5x
    hence integration of e^4ln(x) is = e nd ln cancell each other log rules..
    hence take 4 to the other side becomes power. = x^4
    final ans
    (3x^2 - 5x).(x^4)...so u can do the rest i assume.
    hope been helpful

    to -EquinoX
  9. Sep 25, 2009 #8
    well [tex]e^{4ln(x)}[/tex] is equal to [tex]e^4x[/tex] right?

    if that's so the the integration becomes [tex] e^4 \int (3x^2-5x) [/tex]

    which is [tex](x^3 - \frac{5x^2}{2})e^4 + C[/tex]

    is this correct?

    if so then I divide [tex](x^3 - \frac{5x^2}{2})e^4 + C[/tex] by [tex]e^{4ln(x)}[/tex] and then simplifying that I get [tex] (x^2 - \frac{5x}{2}) + \frac{C}{e^{4ln(x)}} [/tex]
    Last edited: Sep 25, 2009
  10. Sep 25, 2009 #9
    [tex]e^{4 + ln x} = e^{4}e^{ln x} = e^{4}x[/tex]
    [tex]e^{4ln x} = e^{(ln x)4} = (e^{ln x})^{4} = x^{4}[/tex]

    I thought we already went over this. As to the integral question I asked earlier, what's ln 0?
  11. Sep 25, 2009 #10
    shouldn't it be the integral of (3x-5)*x^4\

    I am just confused why you integrate (3x-5) first and then e^4ln(x) separately... as I recall they were multiplied
  12. Sep 25, 2009 #11
    I'm not doing that. I'm finding the integrating factor. That's how you got [tex]e^{4ln x}[/tex], right? By performing the integral I gave above?
  13. Sep 25, 2009 #12
    okay so the integral should be [tex] \int 3x^5-5x^4 = \frac{x^6}{2} - x^5 + C [/tex] then divide all of this by [tex]x^4[/tex] (integrating factor) I will get [tex]\frac{x^2}{2} - x + \frac{C}{x^4} [/tex]

    put in the initial condition which is y(1) = 5,. I can solve for C which is 11/2 then plug it back in I will have [tex]y = \frac{x^2}{2} - x + \frac{11}{2x^4}[/tex] as a particular solution, true or not?
    Last edited: Sep 25, 2009
  14. Sep 25, 2009 #13
    Yes, but in the future, once you have the solution, you can always check it for yourself.
  15. Sep 25, 2009 #14
    and once again reason why I post it is because web assign doesn't accept that answer, I re did the problem couple of times and it goes down to that answer... I hate this thing
  16. Sep 25, 2009 #15
    That's harsh. I hated those things when I had to do them. It's right though. I plugged in for 1 and plugged the solution into the ODE you gave. They're both right so it's not your fault. Maybe put it all in one fraction or something.
  17. Sep 25, 2009 #16
    okay...I'll see what I can do...
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