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#2 2005-10-14 07:39:37
- MathsIsFun
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Re: Integration by substitution
As a start, that should probably be:
∫ x^2/(1 - 2x^3) dx
is that right?
Otherwise it is too easy 
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#3 2005-10-14 13:56:26
- Flowers4Carlos
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Re: Integration by substitution
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 
#4 2005-10-14 15:42:45
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Re: Integration by substitution
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!)
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#6 2005-10-14 17:20:25
- Flowers4Carlos
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Re: Integration by substitution
... to complicate it even further:
ln[1/(1-2x^3)^1/6] + C
#7 2005-10-14 19:11:12
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Re: Integration by substitution
Yes, correct! The constant is necessary for completeness.
∫ x^2/(1 - 2x^3) dx = (-1/6) ln( 1-2x^3 ) + C
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman