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    Integrate: x/x+3 dx, u=x+3

    so du/dx = 1 = dx/du

    Subbing u into the equation:
    x/u

    Rearranging u for x you get u-3=x so therefore (u-3)/u du

    What would the next step be? Because I tried doing (u-3)*u^-1 to get 1-3u^-1
    then integrating that you get u-3lnu which is (x+3)-3ln(x+3)...

    The answer at the back says (x-3)ln(x+3). Any help where I went wrong is greatly appreciated.
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    (Original post by MrToodles4)
    Integrate: x/x+3 dx, u=x+3

    so du/dx = 1 = dx/du

    Subbing u into the equation:
    x/u

    Rearranging u for x you get u-3=x so therefore (u-3)/u du

    What would the next step be? Because I tried doing (u-3)*u^-1 to get 1-3u^-1
    then integrating that you get u-3lnu which is (x+3)-3ln(x+3)...

    The answer at the back says (x-3)ln(x+3). Any help where I went wrong is greatly appreciated.
    You're correct.
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    (Original post by Zacken)
    You're correct.
    Thank you so much. Are you sure? This question really did my head in.
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    (Original post by MrToodles4)
    Integrate: x/x+3 dx, u=x+3

    so du/dx = 1 = dx/du

    Subbing u into the equation:
    x/u

    Rearranging u for x you get u-3=x so therefore (u-3)/u du

    What would the next step be? Because I tried doing (u-3)*u^-1 to get 1-3u^-1
    then integrating that you get u-3lnu which is (x+3)-3ln(x+3)...

    The answer at the back says (x-3)ln(x+3). Any help where I went wrong is greatly appreciated.
    As Zacken said, you are correct. If you're ever in doubt, you can always differentiate again.
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    (Original post by MrToodles4)
    Thank you so much. Are you sure? This question really did my head in.
    Yes. (well you are missing an integration constant + c)

    If the answer at the back of the book is x - 3ln(x+3) + c then this is also correct, the +c absords the +3.

    If the answer at the back of the book is (x-3)ln(x+3) + c as you have written then it is incorrect.
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    (Original post by MrToodles4)
    Thank you so much. Are you sure? This question really did my head in.
    Yes, if you differentiate you have 1 - \frac 3 {x+3} = \frac{x+3 - 3}{x+3} = \frac x {x+3}.
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    (Original post by Zacken)
    Yes. (well you are missing an integration constant + c)

    If the answer at the back of the book is x - 3ln(x+3) + c then this is also correct, the +c absords the +3.

    If the answer at the back of the book is (x-3)ln(x+3) + c as you have written then it is incorrect.
    Oh it does say x-3ln(x+3), my apologies. So is the +3 basically +c? because it did the in the back that the constant is omitted.
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    (Original post by _gcx)
    Yes, if you differentiate you have 1 - \frac 3 {x+3} = \frac{x+3 - 3}{x+3} = \frac x {x+3}.
    How did you get the 1?
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    (Original post by MrToodles4)
    Oh it does say x-3ln(x+3), my apologies. So is the +3 basically +c? because it did the in the back that the constant is omitted.
    Well
    x + 3 - 3 ln(x+3) + c1 and
    x - 3 ln(x+3) + c2

    are both valid integrals, you just have c2 = 3 + c1.
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    (Original post by MrToodles4)
    How did you get the 1?
    d/dx(x+3) = 1
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    (Original post by Zacken)
    Well
    x + 3 - 3 ln(x+3) + c1 and
    x - 3 ln(x+3) + c2

    are both valid integrals, you just have c2 = 3 + c1.
    Oh so both forms are acceptable in the exam?
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    (Original post by MrToodles4)
    How did you get the 1?
    just differentiating x+3.

    As to Zacken's point about arbitrary constant, it doesn't matter if the antiderivatives differ by a constant. If we have (x+3) - 3\ln(x+3) + C_1 and x - 3\ln(x+3) + C_2, we can say C_2 = 3 + C_1, which is what Zacken meant by "absorbed".

    This actually crops up a lot with logs. For example, \displaystyle \int \frac 1 {2x} \mathrm dx. You can extract \frac 1 2 and integrate it to \frac 1 2 \ln(x) + C_1, but equally you can substitute u = 2x to get \frac 1 2 \ln(2x) + C_2. These are equivalent answers, in the second antiderivative, we have \displaystyle \frac 1 2 \ln(2x) + C_2 = \frac 1 2 \ln(2) + \frac 1 2 \ln(x) + C_2. So, we just have C_1 = \frac 1 2 \ln(2) + C_2. Note that if you differentiate both, you'll get the same answer, as the constant will differentiate to zero anyway.
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    (Original post by _gcx)
    Yes, if you differentiate you have 1 - \frac 3 {x+3} = \frac{x+3 - 3}{x+3} = \frac x {x+3}.
    Thank you so much
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    (Original post by MrToodles4)
    Oh so both forms are acceptable in the exam?
    Not really going to dignify that with an answer. Both forms are mathematically correct. This is why we have arbitrary constants when we integrate things, because differentiating constants go to 0 and we use the FToC to integrate things. You should focus on understanding this rather than worry about which forms are acceptable in an exam (they are both correct, so they should both be)
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    (Original post by _gcx)
    just differentiating x+3.

    As to Zacken's point about arbitrary constant, it doesn't matter if the antiderivatives differ by a constant. If we have (x+3) - 3\ln(x+3) + C_1 and x - 3\ln(x+3) + C_2, we can say C_2 = 3 + C_1, which is what Zacken meant by "absorbed".

    This actually crops up a lot with logs. For example, \displaystyle \int \frac 1 {2x} \mathrm dx. You can extract \frac 1 2 and integrate it to \frac 1 2 \ln(x) + C_1, but equally you can substitute u = 2x to get \frac 1 2 \ln(2x) + C_2. These are equivalent answers, in the second antiderivative, we have \displaystyle \frac 1 2 \ln(2x) + C_2 = \frac 1 2 \ln(2) + \frac 1 2 \ln(x) + C_2. So, we just have C_1 = \frac 1 2 \ln(2) + C_2. Note that if you differentiate both, you'll get the same answer, as the constant will differentiate to zero anyway.
    Thanks for this, means a lot - I'd give another rep if I could.
 
 
 
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