Integral
Name:
Anonymous
2008-12-06 9:28
The integral of (1/(sqrt(1+x^3)))dx, entertain me /sci/
Name:
Anonymous
2008-12-06 9:35
Standard integration by substitution. Check your answer using the chain rule.
Name:
Anonymous
2008-12-06 11:17
entertained
Name:
Anonymous
2008-12-06 11:28
No, I'm not.
Name:
Anonymous
2008-12-06 12:07
Simple.
Name:
Anonymous
2008-12-06 13:11
2\, \left( 3/2-1/2\,i\sqrt {3} \right) \sqrt {{\frac {x+1}{3/2-1/2\,i
\sqrt {3}}}}\sqrt {{\frac {x-1/2-1/2\,i\sqrt {3}}{-3/2-1/2\,i\sqrt {3}
}}}\sqrt {{\frac {x-1/2+1/2\,i\sqrt {3}}{-3/2+1/2\,i\sqrt {3}}}}{\it
EllipticF} \left( \sqrt {{\frac {x+1}{3/2-1/2\,i\sqrt {3}}}},\sqrt {{
\frac {-3/2+1/2\,i\sqrt {3}}{-3/2-1/2\,i\sqrt {3}}}} \right) {\frac {1
}{\sqrt {1+{x}^{3}}}}
Name:
Anonymous
2008-12-06 13:13
2\, \left( 3/2-1/2\,i\sqrt {3} \right) \sqrt {{\frac {x+1}{3/2-1/2\,i\sqrt {3}}}}\sqrt {{\frac {x-1/2-1/2\,i\sqrt {3}}{-3/2-1/2\,i\sqrt {3}}}}\sqrt {{\frac {x-1/2+1/2\,i\sqrt {3}}{-3/2+1/2\,i\sqrt {3}}}}{\it EllipticF} \left( \sqrt {{\frac {x+1}{3/2-1/2\,i\sqrt {3}}}},\sqrt {{\frac {-3/2+1/2\,i\sqrt {3}}{-3/2-1/2\,i\sqrt {3}}}} \right) {\frac {1}{\sqrt {1+{x}^{3}}}}
Name:
Anonymous
2008-12-06 18:30
>>6
I think you're making this up
Name:
Anonymous
2008-12-07 11:53
I'm guessing it's -1*(1+x^3)^-1 .
Name:
Anonymous
2008-12-07 16:20
2(sqrt(1+x^3))/(3x^2)
Name:
Anonymous
2008-12-08 3:13
Name:
Anonymous
2008-12-08 9:47
>>11
>Apparently you're all wrong.
Does this suprise you?
Name:
Anonymous
2008-12-08 13:33
>>11
Cheater, do it manually.
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