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    show that integral of 2/x with limits root 6 and root 2 = ln3
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    (Original post by elvss567)
    show that integral of 2/x with limits root 6 and root 2 = ln3
    Go on then.
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    Quite a trivial problem once you: integral of 1/x = ln x (or absolute value for negative values).

    Hence F(root6) - F(root 2) = ln3. You can show this now.
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    (Original post by RDKGames)
    Go on then.
    i integrated 2/x to get 2lnx. then applying the limits gets me:

    (2lnroot 6 ) - (2lnroot2) =

    using log rules i got lnroot 3 as the answer.

    the mark scheme answer says ln3. what have i done wrong when using log rules?
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    (Original post by thekidwhogames)
    Quite a trivial problem once you: integral of 1/x = ln x (or absolute value for negative values).

    Hence F(root6) - F(root 2) = ln3. You can show this now.
    ^^
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    (Original post by elvss567)
    i integrated 2/x to get 2lnx. then applying the limits gets me:

    (2lnroot 6 ) - (2lnroot2) =

    using log rules i got lnroot 3 as the answer.

    the mark scheme answer says ln3. what have i done wrong when using log rules?
    Well 2\ln (\sqrt{6})=\ln(6)=\ln(2)+\ln(3)

    Similarly, try simplifying 2\ln(\sqrt{2})
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    (Original post by elvss567)
    i integrated 2/x to get 2lnx. then applying the limits gets me:

    (2lnroot 6 ) - (2lnroot2) =

    using log rules i got lnroot 3 as the answer.

    the mark scheme answer says ln3. what have i done wrong when using log rules?
    Use logarithm rules properly. 2ln root 6 = ln (root6)^2 = ln6
    Similarly, the second part is ln2.

    Thus ln6-ln2=ln3 as requested.
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    Thankyou both!
 
 
 
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