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Math Help - Integration Question

  1. #1
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    Integration Question

    Hi all,
    Could someone please show me how to integrate:

    [ ( 1 + 4kut ) ^1/2 ] / 2kt dt

    I've tried everything, maybe I'm just being dim but I can't find a substitution that works and I can't do it by parts.
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  2. #2
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    Re: Integration Question

    Quote Originally Posted by Ivanator27 View Post
    Hi all,
    Could someone please show me how to integrate:

    [ ( 1 + 4kut ) ^1/2 ] / 2kt dt

    I've tried everything, maybe I'm just being dim but I can't find a substitution that works and I can't do it by parts.
    \displaystyle \begin{align*} \int{\frac{\sqrt{1 + 4ku\,t}}{2k\,t}\,dt} &= \frac{1}{2k}\int{\frac{\sqrt{1 + 4ku\,t}}{t}\,dt} \\ &= \frac{1}{2k}\int{\frac{1 + 4ku\,t}{t\sqrt{1 + 4ku\,t}}\,dt} \\ &= \frac{1}{4k^2u}\int{\frac{2ku\left(1 + 4ku\,t\right)}{t\sqrt{1 + 4ku\,t}}\,dt} \end{align*}

    Let \displaystyle \begin{align*} v = \sqrt{1 + 4ku\,t} \implies dv = \frac{2ku}{\sqrt{1 + 4ku\,t}}\,dt \end{align*} and the integral becomes

    \displaystyle \begin{align*} \frac{1}{4k^2u}\int{\frac{v^2}{\frac{v^2 - 1}{4ku}}\,dv} &= \frac{1}{4k^2u}\int{\frac{4ku\,v^2}{v^2 - 1}\,dv} \\ &= \frac{1}{k}\int{\frac{v^2}{v^2 - 1}\,dv} \\ &= \frac{1}{k}\int{\frac{v^2 - 1 + 1}{v^2 - 1}\,dv} \\ &= \frac{1}{k}\int{1 + \frac{1}{v^2 - 1}\,dv}\end{align*}

    You can now integrate this using partial fractions
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  3. #3
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    Re: Integration Question



    I agree with that, however once:







    Which leads me to:





    Last edited by Ivanator27; April 26th 2012 at 12:45 AM.
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  4. #4
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    Re: Integration Question

    Let z^2=1+4kut SO 2zdz/dt=4ku So zdz/2ku z^2-1=4kut So 2kt=(z^2-1)/u

    Integral becomes 1/2k integral of z^2/(z^2-1) Now write z^2/(z^2-1)= 1+1/(z^2-1) and partial fraction 1/(z^2-1)
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  5. #5
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    Re: Integration Question

    End of my 1st line should have been 2kt=(z^2-1)/2u
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  6. #6
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    Re: Integration Question





    I don't see how this helps me, it leads to:



    Which is exactly what happens if you let

    Maybe I'm not understanding exactly what you were saying.
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  7. #7
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    Re: Integration Question

    It was just an alternative substitution. You now need to write t in terms of z.
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  8. #8
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    Re: Integration Question

    Ohhhh! Thanks so much! I have it know. =D

    ...Thanks again!
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