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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