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Integrate e^{-2x}.tanh(x)

  1. Apr 15, 2012 #1


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    The problem statement, all variables and given/known data

    Find [tex]\int e^{-2x}\tanh x\,dx[/tex]

    The attempt at a solution

    [tex]\int e^{-2x}\tanh x\,dx
    \\=\int e^{-2x}\times \frac{e^x-e^{-x}}{e^x+e^{-x}}\,dx
    \\=\int \frac{e^{-x}-e^{-3x}}{e^x+e^{-x}}\,dx

    Then i have no idea how to proceed.
  2. jcsd
  3. Apr 15, 2012 #2


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    Hint: can you do anything with partial fractions?
  4. Apr 15, 2012 #3


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    I've already tried to split it:
    [tex]\int \frac{e^{-x}}{e^x+e^{-x}}-\int \frac{e^{-3x}}{e^x+e^{-x}}[/tex]But got stuck again.

    The answer is: [tex]\frac{e^{-2x}}{2}-\ln (1+e^{2x})[/tex]

    I'm thinking maybe integration by parts?
  5. Apr 15, 2012 #4


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    [itex]\displaystyle \frac{e^{-x}}{e^x+e^{-x}}=\frac{e^{-2x}}{1+e^{-2x}}[/itex]

    [itex]\displaystyle \frac{e^{-3x}}{e^x+e^{-x}}=\frac{e^{-4x}}{1+e^{-2x}}\ .[/itex] Then let u = e-x or maybe let u = 1+e-2x .
  6. Apr 16, 2012 #5


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    [tex]\int \frac{e^{-2x}}{1+e^{-2x}}\,.dx-\int \frac{e^{-4x}}{1+e^{-2x}}\,.dx[/tex]
    Let [itex]u =e^{-x}
    \\Then \, \frac{du}{dx}=-e^{-x}=-u[/itex]
    [tex]=\int \frac{-u}{1+u^2}\,.du-\int \frac{-u^3}{1+u^2}\,.du[/tex]
    [tex]=-\frac{1}{2}\ln (1+u^2) + \int u\,.du - \int \frac{u}{(1+u^2)}\,.du[/tex]
    [tex]=-\frac{1}{2}\ln (1+u^2) + \frac{u^2}{2} - \frac{1}{2}\ln (1+u^2)[/tex]
    [tex]=\frac{u^2}{2}-\ln (1+u^2)[/tex]
    Substituting for u, we have the final answer:
    [tex]=\frac{e^{-2x}}{2}-\ln (1+e^{-2x})[/tex]
    But the answer in my copybook is:
    [tex]=\frac{e^{-2x}}{2}-\ln (1+e^{2x})[/tex]
    By the way, isn't there a need to add the arbitrary constant of integration?
    Last edited: Apr 16, 2012
  7. Apr 16, 2012 #6


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    Your result is correct, and add a constant.

  8. Apr 16, 2012 #7


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    Hi ehild!

    Thank you for your confirmation. :smile:
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