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    (u)^(-1/2)dx where u=(5+4x-x^3)
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    (Original post by Whyte-1)
    (u)^(-1/2)dx where u=(5+4x-x^3)
    Hi, I've moved your thread to the maths forum where you'll hopefully receive some help.
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    Thanks a bunch.
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    (Original post by Smack)
    Hi, I've moved your thread to the maths forum where you'll hopefully receive some help.
    Hi is there a link I would go to check that forum.
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    (Original post by Whyte-1)
    (u)^(-1/2)dx where u=(5+4x-x^3)
    I think this is likely to involve elliptic integals. In general you can't integrate square roots of cubics without use of special functions.
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    (Original post by Whyte-1)
    (u)^(-1/2)dx where u=(5+4x-x^3)
    (Original post by RichE)
    I think this is likely to involve elliptic integals. In general you can't integrate square roots of cubics without use of special functions.
    Wolfram gives a page-long expression involving both an elliptic integral and the 3 roots of the polynomial (which are all horrendous expressions in their own right).

    Either the OP has made a mistake, or the integral is supposed to be found numerically.
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    1/sqrt(u)? Think about the relationship between a reciprocal and the logarithm.
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    (Original post by rickyrossman)
    1/sqrt(u)? Think about the relationship between a reciprocal and the logarithm.
    I suggest you explain what you mean; TBH I think you're confused (you've got 2 experienced graduates who don't think this has a solution in terms of elementary functions).
 
 
 
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