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    How do i integrate mv/(-B-kv^2) dv? m B and k are constants
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    (Original post by JordanW)
    How do i integrate mv/(-B-kv^2) dv? m B and k are constants
    \int \dfrac{mv}{-B-kv^2} dv?

    well..

    If m, B and K are constant values:

     \dfrac{-1}{B} \int \dfrac {mv}{-kv^2} dv =



\dfrac{-1}{B} \int \dfrac {m}{-kv} dv =



\dfrac{m}{Bk} \int \dfrac {1}{v} dv.

    Think that's allowed?

    _Kar.
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    (Original post by JordanW)
    How do i integrate mv/(-B-kv^2) dv? m B and k are constants
    If you can't see it directly, use a substitution of u=-(B+kv^2).
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    (Original post by Kareir)
    \int \dfrac{mv}{-B-kv^2} dv?

    well..

    If m, B and K are constant values:

     \dfrac{-1}{B} \int \dfrac {mv}{-kv^2} dv =



\dfrac{-1}{B} \int \dfrac {m}{-kv} dv =



\dfrac{m}{Bk} \int \dfrac {1}{v} dv.

    Think that's allowed?

    _Kar.
    The first step is wrong.
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    As farhan suggested, use that particular substitution and providing you are confident at integration by substitution, it should all come together quite nicely and without too much difficulty.
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    (Original post by Farhan.Hanif93)
    The first step is wrong.
    This is true as B is not a common factor in the denominator
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    (Original post by Farhan.Hanif93)
    The first step is wrong.
    haha, yeah, I noticed just like a few mins after I posted. Sorry.

    _Kar.
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    I dont think substitution works :/
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    Substitution works.
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    Im struggling with this and its apparently pretty simple.
    Im trying to find s using:
    \frac{-B-kv^{2}}{m} = v\frac{dv}{ds}
    hence,
    \int^55_0 \frac{mv}{-B-kv^{2}}\ dv = \int 1\ ds

    The integration should be between 55 and 0 but i have only just started using latex and cant seem to get it to work. I know what i have done thus far is correct.
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    As far as the latex goes, you need to put 55 in braces: {55} if you want it to all be superscripted.

    As far as the integral goes, use the substitution Farhan suggested.
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    Cheers for the help guys, i think it should be:
    \frac{m}{-2k}ln(B+kv^2)
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    (Original post by JordanW)
    Cheers for the help guys, i think it should be:
    \frac{m}{-2k}ln(B+kv^2)
    That's correct.
 
 
 
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