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A problematic integral

  1. Feb 17, 2007 #1
    i added a file with with this integral

    in wich i tried to solve it

    how do i solve this case?

    plz help
     

    Attached Files:

  2. jcsd
  3. Feb 17, 2007 #2
    If the top is a derivative of the bottom in this case:-

    [tex]\frac{e^{2x}-1}{e^{2x}+3}[/tex]

    derivative of [tex]\frac{d}{dx}e^{2x}-1=2e^{2x}[/tex]

    using the log rule for integrals.

    [tex]h(x)=\frac{derivative of bottom}{bottom}=ln(bottom)[/tex]

    Since it is a derivative by a factor of

    [tex](e^{2x}+3)^{2/3}=ln(e^{2x}+3)^{2/3}[/tex]

    I think that's what they've done for d/dt.

    [tex]\frac {2(e^{2x})}{e^{2x}+3}[/tex]
     
    Last edited: Feb 17, 2007
  4. Feb 17, 2007 #3
    if what you say is true than the
    question must be constructed otherwise
    because the derivative of the bottom is NOT what is on TOP??

    i saparated them into 2 integrals the second part is the
    problematic
    because when
    i take the whole
    1/(e^2x+3) and i and take this as a complex function LAN (as shown in the file)
    ln(e^2x+3)/(2*e^2x) it doesnt come out as in the book

    what is the law of transforming into lan fuction exept of the case 1/x

    i meen in the case of complex function
    when i have 1/(something else then simple lenear stuff)????/
     
  5. Feb 17, 2007 #4
    I think your right looking at the answer in d/dx it looks like this

    [tex]\frac {2}{3}ln(3+e^{2x})-\frac{1}{6}ln(e^{2x})[/tex]

    A bit different.

    and for [tex]\frac{d}{dt}[/tex] the paremetric, I get nothing like your books answer either? Sorry I couldn't be of more help, I don't know what they've done then? I'm sure one of the resident experts can figure it out?

    [tex] \frac{e^{2x}+1}{e^{2x}-3}=\frac{d}{dx} of bottom=2e^{2x}[/tex]

    which is

    [tex]\frac {1}{2}(e^{2x})[/tex]

    or [tex]\frac{1}{2}(top)[/tex]

    was what I was thinking, but this is apparently not right.
     
    Last edited: Feb 17, 2007
  6. Feb 17, 2007 #5
    how i deside if its this is lan function

    i can take every time all the function on the buttom
    put it in a lan
    an devide it in its derivative?

    even if i got a fuction
    1/(x+2)^2

    its integral solves [(x+2)^-1]/-1

    or i can open this equetion (a+b)^2=x^2+2*x*2+2^2
    then put it all in a lan ln(x^2+2*x*2+2^2) and i will devide it by it derivative

    2x+4

    ln(x^2+2*x*2+2^2)/2x+4

    what i need to do?
    can some one make reason in this problem
     
  7. Feb 18, 2007 #6
    try solving with the inverse tangent function.

    1/[3((e^t/srt3)^2+1)] , try integrating that and tell me if you get what you are looking for
     
  8. Feb 18, 2007 #7

    arildno

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    It is perhaps simplest to use the variable change [itex]u=e^{x}\to{du}=\frac{du}{u}[/itex]

    Then we get:
    [tex]\int\frac{e^{2x}-1}{e^{2x}+3}dx=\int\frac{u^{2}-1}{(u^{2}+3)u}du[/tex]
    We then use partial fractions decomposition:
    [tex]\frac{u^{2}-1}{(u^{2}+3)u}=\frac{Au+B}{u^{2}+3}+\frac{C}{u}\to{C}=-\frac{1}{3}, A=\frac{4}{3},B=0[/tex]
    Then your problem is readily solved
     
  9. Feb 18, 2007 #8
    thank u very much
     
  10. Feb 19, 2007 #9

    D H

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    Arildno did fine. He did not solve the problem in full; he left the final steps up to the original poster.

    The only problem with Arildno's post is the use of the word "simplest":
    It is even simpler to use the variable change
    [tex]u=e^{2x}\to{dx}=\frac1 {2u}du}[/tex]

    Then
    [tex]\int\frac{e^{2x}-1}{e^{2x}+3}dx\to
    \int\frac1 2\; \frac{u-1}{u(u+3)}du[/tex]
    Decomposing,
    [tex]\frac1 2\frac{u-1}{u(u+3)} = \frac1 6\left(\frac4{u+3}-\frac1u\right)[/tex]
    which leads to the easily integrable function
    [tex]\int\frac1 6\left(\frac4{u+3}-\frac1u\right)du[/tex]
     
  11. Feb 19, 2007 #10
    wrong, on the demominator, you cannot factor out u from 3, a constant. look carefully...
     
  12. Feb 19, 2007 #11

    cristo

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    Which line do you think is wrong?
     
  13. Feb 19, 2007 #12

    D H

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    What are you talking about? The technique is called "partial fraction decomposition". A reference: http://mathworld.wolfram.com/PartialFractionDecomposition.html

    It is easy to verify that
    [tex]\frac1 2\frac{u-1}{u(u+3)} = \frac1 6\left(\frac4{u+3}-\frac1u\right)[/tex]
    The least common denominator of the sum on the left hand side is [itex]u(u+3)[/itex]. Expanding the left hand side,
    [tex]\frac1 6\left(\frac4{u+3}-\frac1u\right) =
    \frac1 6\left(\frac4{u+3}-\frac1u\right)\frac{u(u+3)}{u(u+3)} =
    \frac1 6\;\frac{4u-(u+3)}{u(u+3)} = \frac1 6\;\frac{3u-3}{u(u+3)} = \frac1 2\frac{u-1}{u(u+3)}[/tex]
    which is of course the right-hand side.
     
  14. Feb 19, 2007 #13
    what I am trying to say is that the substitution does not equate to the original equation if you observe the denominator, it is not hard to spot, very obvious blunder...
     
  15. Feb 19, 2007 #14

    D H

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    Point out this "obvious blunder". The only "obvious blunder" is that you apparently do not know how to use substitution to make integration easier.

    Since the OP is long gone, I will finish the integration in post #10.
    [tex]\int\frac1 6\left(\frac4{u+3}-\frac1u\right)du =
    \frac 1 6\left(4\log(u+3)-\log(u)\right) + C[/tex]
    This result is in terms of [itex]u[/itex]. The desired result is in terms of the original variable [itex]x[/itex]. Applying the substitution [itex]u=e^{2x}[/itex],
    [tex]\int\frac{e^{2x}-1}{e^{2x}+3}dx =
    \frac 1 6\left(4\log(e^{2x}+3)-2x\right) + C =
    \frac 1 3\left(2\log(e^{2x}+3)-x\right) + C[/tex]

    To verify, differentiate the right-hand side wrt [itex]x[/itex]:
    [tex]\frac d{dx} \left(\frac 1 3\left(2\log(e^{2x}+3)-x\right)+C\right) =
    \frac 1 3\left(2\frac{2e^{2x}}{e^{2x}+3} - 1\right)[/tex]
    Simplifying,
    [tex]\frac 1 3\left(2\frac{2e^{2x}}{e^{2x}+3} - 1\right) =
    \frac 1 3\;\frac{4e^{2x}-(e^{2x}+3)}{e^{2x}+3} =
    \frac{e^{2x}-1}{e^{2x}+3}[/tex]
    which of course is the function to be integrated.
     
  16. Feb 20, 2007 #15

    arildno

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    We have, by the values for A, B, C:
    [tex]\frac{\frac{4}{3}u+0}{u^{2}+3}+\frac{\frac{-1}{3}}{u}=\frac{(\frac{4}{3}u+0)u+(-\frac{1}{3})(u^{2}+3)}{(u^{2}+3)u}=\frac{(\frac{4}{3}-\frac{1}{3})u^{2}+0u+(-\frac{1}{3})*3}{(u^{2}+3)u}=\frac{u^{2}-1}{(u^{2}+3)u}[/tex]
    as we should have.

    In the future, mathgician, please remain silent when you haven't got the competence to make an informed contribution

    And, DH:
    I DID qualify "simplest" by the word "perhaps"..:smile:
     
    Last edited: Feb 20, 2007
  17. Feb 21, 2007 #16

    dextercioby

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    Here's my treat

    [tex] \int \frac{e^{2x}-1}{e^{2x}+3}{}dx =\int \frac{e^{2x}+3-4}{e^{2x}+3}{}dx =x-4\int \frac{e^{-2x}}{1+3e^{-2x}}{}{dx}=x+\frac{2}{3}\int\frac{d\left(1+3e^{-2x}\right)}{1+3e^{-2x}}[/tex]

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

    Simpler or simplest ? :biggrin:
     
  18. Feb 21, 2007 #17

    arildno

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    Straightforward, but not necessarily simplest. :smile:
     
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