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    Hi I was wondering if you had to integrate 3x squared / 1-2x cubed how would you do it?

    I known that it would be using ln but so it would be ln(1-2x cubed) but what happens to the top part of the fraction when integrating?
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    substitution or partial fractions
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    if it is

    3x2
    ----------
    1 - x3

    then the top is nearly the derivative of the bottom so ln is the way to go...
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    (Original post by the bear)
    if it is

    3x2
    ----------
    1 - x3

    then the top is nearly the derivative of the bottom so ln is the way to go...
    Hi, I know that it would be ln(1-x cubed) but what would happen to the 3x2
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    (Original post by Teddysmith123)
    Hi, I know that it would be ln(1-x cubed) but what would happen to the 3x2
    if you have to integrate

    f'(x)
    ------
    f(x)

    the answer is just ln|f(x)| + c

    the top does not reappear.

    however in this case the top is not the exact derivative of the bottom...
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    (Original post by Teddysmith123)
    Hi, I know that it would be ln(1-x cubed) but what would happen to the 3x2
    It wouldn't be \ln|1-x^3| though that is very close


    Differentiate \ln|1-x^3| and the answer to your question should be obvious
 
 
 
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