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Topic review (newest first)
- MathsIsFun
- 2005-10-14 19:11:12
Yes, correct! The constant is necessary for completeness.
∫ x^2/(1 - 2x^3) dx = (-1/6) ln( 1-2x^3 ) + C
- Flowers4Carlos
- 2005-10-14 17:20:25
... to complicate it even further:
ln[1/(1-2x^3)^1/6] + C
- ganesh
- 2005-10-14 17:02:47
A little further.....
(-1/6) ln( 1-2x^3 )=
ln[1/(1-2x^3)^1/6]
MathsIsFun had done it right, I posted this
.... just to complicate things 
- MathsIsFun
- 2005-10-14 15:42:45
Ouch, that is a tough one!
But I didn't get around to solving "Student"s integral. Which I will attempt now ...
∫ x^2/(1 - 2x^3) dx
Try: u=1-2x^3
Then: du = -6x^2 dx (ie we differentiated it)
So: x^2 dx = (-1/6) du
Substituting 1-2x^3=u and x^2 dx = (-1/6) du we get:
∫ x^2/(1 - 2x^3) dx = ∫ 1/u (-1/6) du
Then it is fairly easy to solve ∫ 1/u (-1/6) du:
∫ 1/u (-1/6) du = (-1/6) ∫ 1/u du = (-1/6) ln u
Now put back u=1-2x^3:
(-1/6) ln( 1-2x^3 )
DONE! (I hope!)
- Flowers4Carlos
- 2005-10-14 13:56:26
if u think that is too easy try this one:
∫(x²-29x+5)/[(x-4)²(x²+3)]dx
this tuff question was on my calculus2 final, and no, i did not solve it 
- MathsIsFun
- 2005-10-14 07:39:37
As a start, that should probably be:
∫ x^2/(1 - 2x^3) dx
is that right?
Otherwise it is too easy 
- Student
- 2005-10-14 01:40:07
Can somebody please help me!
∫ x^2/1 - 2x^3