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Math Help - Integration question that I 'should' find easy

  1. #1
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    Integration question that I 'should' find easy

    I was recently doing a past paper and I got myself in a muddle when trying to integrate 2/(2x + 1).

    Basically I was just wondering why the integral of the above problem is:
    ln(2x + 1) as opposed to 2ln(2x + 1).

    I would really appreciate a detailed answer. Thanx
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  2. #2
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    \int 2/(2x+1) dx = 2 * \int 1/(2x+1)dx

    Substitute 2x+1 =: z, then

    2 \int 1/(2z) dz = \int 1/z dz = ln(z) = ln(2x+1)

    Done
    Last edited by Rapha; May 29th 2009 at 09:19 AM. Reason: I made a mistake... Thanks to HallsofIvy
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  3. #3
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    Thanx, so im guessing the markscheme was wrong to say the integral was just ln(2x+1)?
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  4. #4
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    Quote Originally Posted by Rapha View Post
    \int 2/(2x+1) dx = 2 * \int 1/(2x+1)dx

    Substitute 2x+1 =: z, then

    2 \int 1/z dz = 2*ln(z) = 2*ln(2x+1)

    Done
    No. If z= 2x+ 1 then \frac{dz}{dx}= 2 so dz= 2dx.

    2\int \frac{dx}{2x+1}= \int \frac{2dx}{2x+1}= \int \frac{dz}{z}
    ln|z|+ C= ln|2x+1|+ C
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