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Why do you use modulus with ln when integrating? C4 Watch

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    Can someone please explain why we use the modulus in the ln when integrating. The book says that the reason for this is explained in 6.5 however it doesn't say anything.

    THanks mike
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    who is mike?
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    Because the natural log of non-positive numbers is not defined, and ln|x| = lnx if x > 0, with derivative 1/x, but ln|x| = ln(-x) if x < 0, with derivative -1(1/-x) = 1/x
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    (Original post by michael242103)
    Can someone please explain why we use the modulus in the ln when integrating. The book says that the reason for this is explained in 6.5 however it doesn't say anything.

    THanks mike
    (Original post by 13 1 20 8 42)
    Because the natural log of non-positive numbers is not defined, and ln|x| = lnx if x > 0, with derivative 1/x, but ln|x| = ln(-x) if x < 0, with derivative -1(1/-x) = 1/x
    As he said, the natural log of non-positive numbers is not defined. You might still question why it's okay to just throw a modulus on it though.

    One reason is because ln(-x) + c= ln(-1) + ln x + c = ln x + k where k = c + ln(-1) since ln(-1) is just a constant. (it's actually ipi). So in the end you get ln x anyway. That justifies you throwing on the modulus signs.
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    (Original post by TeeEm)
    who is mike?
    me mate
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    (Original post by GeologyMaths)
    me mate
    were you not banned? (again?)
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    (Original post by TeeEm)
    were you not banned? (again?)
    dont even start mate
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    so try  \dfrac{d}{dx}lnx and draw the graph baby
 
 
 
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