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Help with integral

  1. May 7, 2007 #1
    Hey, just did a test and have no clue about this question:


    e^(x)/(e^(2X) + 9)

    Really curious how this is done. Thanx
  2. jcsd
  3. May 7, 2007 #2


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    Well, the numerator is (a constant times) the derivative of the denominator. Can you integrate a function of the form [tex]\frac{f'(x)}{f(x)}[/tex]?
  4. May 7, 2007 #3
    I thought that too at first, but then realized that the numerator is e^(x) and not e^(2x), which would have made it a lot simpler. had it been e^(2x) i could have used ln(f(x))
  5. May 7, 2007 #4
    Do a change of variables e^x to z, this yields a standard integral of the arctangent.
  6. May 7, 2007 #5


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    Sorry, I misread that! Anyway, willem's idea seems the way to go!
  7. May 7, 2007 #6
    Yeah i tried that on the test, but didnt get too far, probably cause i havnt worked too much with arctangents (arcsins or arccosins for that matter) in calculus.

    I prety much put down (1/3)arctan(e^(x)/3), but it gave me the wrong answer (PS: it was the finite integral from 0 to ln(3), so i could test it on my calculator)
  8. May 7, 2007 #7

    Gib Z

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    Well [tex]\int \frac{e^x}{(e^x)^2 +9} dx[/tex] is what makes the substitution u=e^x easier to see. du = e^x dx

    That makes the integral [tex]\int \frac{1}{u^2+9} du[/tex] which is of the arctan form, but if you haven't learned that then use the substitution u= 3 tan theta.
  9. May 8, 2007 #8
    Well my formula booklet says: integral of 1/(a^2+x^2) = (1/a)arctan(x/a) + c

    Using that logic and substitution, i got: (1/3)arctan(e^(x)/3), which isnt right. what did i do wrong here?

    Using e^x =3tan(theta) im left with dx= 3sec^2(theta) d(theta), which turns out ugly
  10. May 8, 2007 #9

    Gib Z

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    Forget about replacing u with e^x with respect to my suggested substitution for now.

    [tex]\int \frac{1}{u^2+3^2} du = \frac{1}{3} \int \frac{1}{\sec^2 \theta} \cdot \sec^2 \theta d\theta = \frac{1}{3} \theta + C = \frac{1}{3} \arctan (u/3) + C = \frac{1}{3} \arctan (\frac{e^x}{3}) +C[/tex] which turns out the same as your formula booklet! thats because for solving that general integral, we use u=a tan theta!

    Why in the world do you think thats not correct!!
    Last edited: May 8, 2007
  11. May 8, 2007 #10
    for two reasons, firstly because calculating the finite ingtegral from 0 to ln(3) i get a different answer than using that formula (TI-83 plus). Secondly, when graphing that equation and the original equation it doesnt look like the first derivative to me.

    Maybe im just screwed up and tried everything in the wrong mode or something. Using that formula, the answer will be in radians right?

    Thanx for your time btw
  12. May 8, 2007 #11

    Gib Z

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    When EVER we do calculus, everything is in Radians thanks :P

    Now the answer I get is [itex]\pi/12 - 1/3 \arctan (1/3)[/itex], which turns out around 0.154549203. If that TI 83 plus is different, then I can finally be assured i know more than a calculator.
  13. May 8, 2007 #12


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    foges, when you say taking the finite integral from 0 to ln(3), you mean you take the integral from 0 to ln(3) of the original integral in your first post, then compare it to

    [tex]\frac{1}{3}*(artcan(e^{ln(3)}/3) - arctan(e^0/3))[/tex]
  14. May 8, 2007 #13
    Ok, just re did it on my calculator and i got that answer, so strange, maybe i subtracted arctan(0) or had it in the wrong mode or something. Anyways, thanx for the help.
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