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integrating 1/2x Watch

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    how come when i integrate it by doing 0.5 x integral of 1/x

    i get a different answer to just integrating it without taking the factor of a half,

    so for the first one i get 1/2 ln(x)

    and for the second one i get 1/2 ln (2x)

    (i know when i differentiate both of them, i get the same result, but which is the correct integral?)
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    Both are but the C values will be different.
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    (Original post by cvat)
    how come when i integrate it by doing 0.5 x integral of 1/x

    i get a different answer to just integrating it without taking the factor of a half,

    so for the first one i get 1/2 ln(x)

    and for the second one i get 1/2 ln (2x)

    (i know when i differentiate both of them, i get the same result, but which is the correct integral?)
    They are both correct. The only difference is that in each case, the constant of integration will be different. Obviously this doesn't matter if you're substituting in limits, as the constant cancels out anyway - you'll get the same answer for both.
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    its 1/2 ln x
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    (Original post by cvat)
    how come when i integrate it by doing 0.5 x integral of 1/x

    i get a different answer to just integrating it without taking the factor of a half,

    so for the first one i get 1/2 ln(x)

    and for the second one i get 1/2 ln (2x)

    (i know when i differentiate both of them, i get the same result, but which is the correct integral?)
    As the above poster said, both are correct, but as a general rule, if have an integral \displaystyle\int \frac{1}{kx} \; dx I find that you should just leave it as \frac{1}{k} \ln x + C. But that's just me...
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    so if it wasnt a definite integral, and it asked me that question, i'd get the marks for either of them?
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    (Original post by cvat)
    so if it wasnt a definite integral, and it asked me that question, i'd get the marks for either of them?
    Yes.
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    ok thanks , but yh i think 1/2 ln x + C, does look better too
 
 
 
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