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Integration of e powers

  1. Sep 21, 2004 #1

    Pythagorean

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

    I tried factoring the bottom to ([tex]{e^x+2}[/tex])([tex]{e^x+1}[/tex]) and using PFDs

    and I've also tried [tex] u=2x [/tex] and [tex] u=e^x [/tex]

    We haven't covered e operations in class and th book gives no examples. I assumed it would just be simple enough to do (A + B)e^x.

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

    I've filled up six pages with this problem, and i'm so close. is there something I'm forgetting from back in the day?
     
    Last edited: Sep 21, 2004
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  3. Sep 21, 2004 #2

    Pythagorean

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    i've also tried long division, but that yields a 1, which integrates into an X, which does not comply with the answer :/
     
  4. Sep 21, 2004 #3
    let t=e^x, means 1-(3t+2)/(t^2+3t+2), let u=3t+2, take dt/du, then t^2=(u-2)^2/9, substitute this into the previous integrte, bring the denominator of the just-obtained integration to the form 1/(a^2+ or - b^2) and use known formula of this integration to get the wanted result, of course there are still some steps to make things neater. that is where i have come, have no time to go past the rest, you have to do that homework yourself. and hope this helps.
     
    Last edited: Sep 21, 2004
  5. Sep 21, 2004 #4

    HallsofIvy

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    Kroneker gave a good answer but I'll say my own piece: you said you tried u= ex. If you did then you found that du= exdx. You may have decided that that wouldn't work because you didn't have an "ex". Actually, you do! You have e2x= (ex)2= exex.
    Letting u= ex the numerator is exexdx= u du while the denominator becomes u2+ 3u+ 2= (u+2)(u+1). The integral becomes
    integral of udu/((u+2)(u+1)). That can be done by "partial fractions".
    u/((u+2)(u+1))= A/(u+2)+ B/(u+1). Multiplying both sides by (u+2)(u+1), we get
    u= A(u+1)+ B(u+2). Taking u= -1, that is -1= B. Taking u= -2, -2= -A so A= 1.
    The integral is the same as 1/(u+2)- 1/(u+1). Can you integrat that?
     
  6. Sep 21, 2004 #5

    Pythagorean

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    wow, man, I finally figured this one out. Thank you everyone for your help. I actually solved this using a book in the reference section of the library to get how to integrate e^x.

    The processes are a) long division, b) partial fraction decomposition (as suggest by Halls of Ivy), c) integration of terms, and finally d) arrangement of natural logs

    it came out to be something like x-2x+2ln (x^2+2) + x - ln (x^2+1)

    which simplifies to 2ln (x^2+2) - ln (x^2+1) which yields the answer from the first post when arranged properly for logs
     
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