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**Student****Guest**

Can somebody please help me!

∫ x^2/1 - 2x^3

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

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

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**Flowers4Carlos****Member**- Registered: 2005-08-25
- Posts: 106

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

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

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

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,331

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

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**Flowers4Carlos****Member**- Registered: 2005-08-25
- Posts: 106

... to complicate it even further:

ln[1/(1-2x^3)^1/6] + C

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

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

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