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    If I were to integrate x+2/x-1 let u = x-1
    du/dx=1 = dx/du

    (x+2)/u du and rearranging u for x gives u+1=x

    So substituting x in we get (u+3)/u

    which simplifies to 1+3/u integrating this gives me u+3lnu which is (x-1)+3ln(x-1). where did i go wrong? because the answer at the back says x+3ln(x-1)

    All help is greatly appreciated thanks.
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    This is basically the same as the last thread. Remember your constants of integration.

    You have (x-1) + 3\ln(x-1) + C_1 and x+3\ln(x-1) + C_2, differing by only a constant. We have C_1 - 1 = C_2. This constant will differentiate to zero so both of these functions will have the same derivative, hence are both valid antiderivatives. No need to have multiple threads on the same thing
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    (Original post by _gcx)
    This is basically the same as the last thread. Remember your constants of integration.

    You have (x-1) + 3\ln(x-1) + C_1 and x+3\ln(x-1) + C_2, differing by only a constant. We have C_1 - 1 = C_2. This constant will differentiate to zero so both of these functions will have the same derivative, hence are both valid antiderivatives. No need to have multiple threads on the same thing
    Ohhh yes this is exactly the same case! I completely forgot - thanks again
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    I believe the method the book used was polynomial division on the fraction - which makes sense if this were c3, and I would do it anyway if possible for c4.

    So (x+2)/(x-1) = ((x-1)+3)/(x-1) = 1 + 3/(x-1)
    Which clearly integrates to the book answer.
    Of course as gcx said they are both valid answers, this is just a much simpler method.
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    (Original post by carpetguy)
    I believe the method the book used was polynomial division on the fraction - which makes sense if this were c3, and I would do it anyway if possible for c4.

    So (x+2)/(x-1) = ((x-1)+3)/(x-1) = 1 + 3/(x-1)
    Which clearly integrates to the book answer.
    Of course as gcx said they are both valid answers, this is just a much simpler method.
    I haven't been taught this, so is the exact method to always just get ((x-1)+3) to match the denominator? But I see, thank you And also what type of questions would I be able to use this method for, does i work for all? And this integrates to the answer I got - not the book answer.
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    (Original post by MrToodles4)
    I haven't been taught this, so is the exact method to always just get ((x-1)+3) to match the denominator? But I see, thank you And also what type of questions would I be able to use this method for, does i work for all? And this integrates to the answer I got - not the book answer.
    Well what I did was a specialised form of polynomial division.

    x+2 ÷ x-1 = 1 remainder 3.

    So x+2 = 1×(x-1) + 3

    Then I split up the fraction.

    It should work for pretty much everything, except when the power on the top is higher in which case this precedes partial fractions.
    Someone can probably give you a better idea of when to use it than me since I'll just use it if I can - I don't follow any "set rules".

    But aside from integration this is handy in curve sketching and differentiation too. It really is just polynomial division.


    Btw, it should integrate to the book answer.
 
 
 
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