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Homework Help: Find the integral

  1. Jun 28, 2008 #1

    [tex]\int[/tex] (e[tex]^{}x[/tex] + 1)/e[tex]^{}x[/tex] dx

    2. Relevant equations

    the answer is: x-e[tex]^{}-x[/tex] + C
    after I integrate I don't get the above answer.

    3. The attempt at a solution
    =[tex]\int[/tex] u du
  2. jcsd
  3. Jun 28, 2008 #2


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    Well your method is wrong. If you let u = e^x + 1, du = e^xdx, then you have udu = (e^x+1)(e^xdx), which is clearly different from (e^x+1)/(e^x).

    Hint: (a+b)/a = 1 + b/a.
  4. Jun 28, 2008 #3
    Thanks for your help.

    [tex]\int[/tex] u/du
    I don't know how to start integrating this fraction (that's if this is correct).
    Integrate the top (u) and then the bottom (du)?
  5. Jun 28, 2008 #4
    Is this correct?
    [tex]\int[/tex] U/du=
    e[tex]^{}x[/tex] /e[tex]^{}x[/tex] + 1
    =1/e[tex]^{}x[/tex] + X
    =e[tex]^{}-x[/tex] - X
  6. Jun 28, 2008 #5
    Why did it become a negative x?

  7. Jun 28, 2008 #6
    I'm sorry, I should have typed:
    =1/(e[tex]^{}x[/tex] + x)
    and when I brought it up to get rid of the fraction, then it became (e[tex]^{}-x[/tex] -x
  8. Jun 28, 2008 #7
    You're breaking up the fraction incorrectly.
  9. Jun 28, 2008 #8
    ??? confused, I will review the problem again. Any suggestions?
  10. Jun 28, 2008 #9
    [tex]\int(e^{-x}+1)dx=\pm e^{-x}\pm x+C[/tex]

    What should be the proper signs? positive or negative for e^(-x) and x ???

    What's incorrect with the arithmetic is that ... [tex]\frac{1}{e^x+x}\neq\(e^{-x}-x[/tex] ... it would be [tex]\frac{1}{e^x+x}=(e^{x}+x)^{-1}[/tex]
    Last edited: Jun 28, 2008
  11. Jun 29, 2008 #10


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    One other problem with this is that du can never be in the denominator in an integration. Keep in mind that integration is the inverse of differentiation and that

    [tex] F(x) = \int F'(x) dx = \int \frac{dF}{dx} dx = \int dF [/tex]

    dF being the "differential" of F. Having the differential dx in the denominator does not make sense in terms of the meaning of the symbols for integration.

    As for the rest of post #4, I think you were on the right track, but then confused matters. It looks like you were finding the quotient and integrating it at the same time. Maybe we should go back and ask first: what does [tex]\frac{e^{x} + 1}{e^x}[/tex] simplify to?

    As another question relevant to the solution, what is the anti-derivative of [tex]e^{-x}[/tex]?
    Last edited: Jun 29, 2008
  12. Jun 29, 2008 #11
    Looks like I'm an idiot,:frown: but I'm trying to understand this "integration" concept. I appreciate your help:smile:.

    simplified to: 1+(1/e^x)
    integrating above= x-(e^-x)
    seems like the answer in the solutions manual:

    Is this process correct?
  13. Jun 29, 2008 #12
    Ok, so who cares about the answer in the SM. What is YOUR answer? :)

    And, yes that is the correct simplification.
  14. Jun 29, 2008 #13

    my answer is:
    x+(-e^-x) + C
  15. Jun 29, 2008 #14


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    There you go! And you can always check the result of an integration (not always easy, but particularly important when your SM doesn't include an answer) by differentiating it:

    d/dx [ x - (e^-x) + C]

    = 1 - [ (e^-x)·{ d/dx (-x) } ] + 0

    = 1 - (-1)·(e^-x) = 1 + e^-x ,

    or, multiplying by (e^x)/(e^x) ,

    [ { 1 + (e^-x) } · e^x ] / (e^x) = ( {e^x} + 1 ) / (e^x) ,

    your original integrand.

    And, no, you're not "an idiot". Just review some of your algebra and the definition of integration.... and keep practicing!
    Last edited: Jun 29, 2008

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