integration of sqrt of a cubic functionsWatch

#1
(u)^(-1/2)dx where u=(5+4x-x^3)
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8 months ago
#2
(Original post by Whyte-1)
(u)^(-1/2)dx where u=(5+4x-x^3)
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#3
Thanks a bunch.
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#4
(Original post by Smack)
Hi is there a link I would go to check that forum.
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8 months ago
#5
(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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8 months ago
#6
(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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8 months ago
#7
1/sqrt(u)? Think about the relationship between a reciprocal and the logarithm.
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8 months ago
#8
(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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