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Indefinite integral

  1. May 3, 2009 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    answer: -arctan[5/exp(6*x)]/30

    3. The attempt at a solution

    honestly, don't know where to start. i was looking at another forum and tried to set u=exp(x) du=exp(x) and dx=du/u. plugging that in i got u^6/(u^12+25)*du/u. not sure where to go from there or if that is even the way to go.
  2. jcsd
  3. May 3, 2009 #2


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    Gold Member

    Hi supraboy, Welcome to PF!:smile:

    Try the substitution [tex]u=e^{6x}[/tex] instead :wink:
  4. May 3, 2009 #3
    Try setting u = e6x. Then du = 6e6x and e12x = u2.
  5. May 3, 2009 #4
    ok, setting u=e^6x du=6e^6x, then dx=du/6u?

    then, it would be int(u/u^2+25)du

    using the formula int(a^2+u^2) = (1/a)arctan(u/a) + C

    i get, (1/5)arctan(e^6x/5)dx or (1/30)arctan(e^6x/5) + C

    this is incorrect though because the answer is negative and it should be arctan(5/e^6x) instead of arctan(e^6x/5).

    any ideas?
    Last edited: May 3, 2009
  6. May 4, 2009 #5
    You made a mistake with the substitution. Write your integral like this:
    [tex] \int \frac{(e^{6x} dx)}{(e^{6x})^2 + 25} [/tex]
    If [tex] u = e^{6x} [/tex], then [tex] e^{6x} dx = \frac{1}{6} du [/tex]. Try working from there.
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