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Integral of this simple function

  1. Jul 4, 2013 #1
    im trying to find...

    ∫te^(5t)dt

    i did integration by parts..

    u=e^5t
    dv=tdt

    ∫te^(5t)dt={e^(5t)}(t)-5∫(t){e^(5t)}dt

    collected the integral terms on one side to get..

    6∫te^(5t)dt={e^(5t)}(t)

    ∫te^(5t)dt= {e^(5t)}(t)/6

    but wolfram alpha says otherwise. they don't give out the steps freely anymore. please help
     
  2. jcsd
  3. Jul 4, 2013 #2

    Ackbach

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    Use ILATE for integration by parts:

    Inverse trig
    Logarithmic
    Algebraic
    Trig
    Exponential

    Whichever function is higher on the list gets to be u, and dv is whichever function is lower. You have a t, which is algebraic, and you have an exp(5t), which is exponential. Try that.
     
  4. Jul 4, 2013 #3

    Ray Vickson

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    With u = e^(5t) and dv = t dt, you will get

    ∫ t e^(5t) dt = (1/2)t^2 e^(5t) - (5/2) ∫ t^2 e^(5t) dt,

    so you have just made it worse! You have the wrong u and dv.
     
  5. Jul 5, 2013 #4

    vanhees71

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    Define the function
    [tex]f(t,x)=\int \mathrm{d} t \exp(t x)=\frac{1}{x} \exp(t x).[/tex]
    Then you have
    [tex]\partial_x f(t,x)=\int \mathrm{d} t \exp(t x)=-\frac{1}{x^2} \exp(t x)+\frac{t}{x} \exp(t x).[/tex]
    Setting [itex]x=5[/itex] you find
    [tex]\int \mathrm{d} t \exp(5t)=\exp(5t) \left [\frac{t}{5}-\frac{1}{25} \right].[/tex]
     
  6. Jul 8, 2013 #5
    There is a bit of faulty notation here. After partially differentiating f with respect to t, you should have an additional t factor inside the integral, which is what I think you wanted, as your solution seems to be using that. I believe you have forgot to type it..?
     
  7. Jul 8, 2013 #6

    CAF123

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    I never memorised this list - I just look at what could possibly be u and dv and then decide what would ultimately simplify the integral. As Ray said, by choosing dv = t dt, all you do is introduce greater powers of t whereas choosing u = t, when you differentiate it, you simply get 1 and hence is more likely to simply the integral. I think (i may be wrong) this is how that ILATE list was created - after some experience, it was put together as an easy way to decide on u and dv which would lead to a simpler integral.
     
  8. Jul 8, 2013 #7

    vanhees71

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    Sure, that's a typo :-(. Correctly it should read
    [tex]\partial_x f(t,x)=\int \mathrm{d} t \; t \exp(t x)=\ldots[/tex]
     
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