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    • Thread Starter
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    Hi so I'm working through past papers and came across this...

    

\int-\dfrac{3}{2x},dx\

    I thought the answer was...

    

-\dfrac{3}{2} \ln |2x| +c

    But the mark scheme has...

    

-\dfrac{3}{2} \ln |x| +c

    Can someone explain this please? It's for CCEA C3.

    Thanks
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    (Original post by 01Chris02)
    Hi so I'm working through past papers and came across this...

    

\int-\dfrac{3}{2x},dx\

    I thought the answer was...

    

-\dfrac{3}{2} \ln |2x| +c

    But the mark scheme has...

    

-\dfrac{3}{2} \ln |2| +c

    Can someone explain this please? It's for CCEA C3.

    Thanks
    The mark scheme is wrong.
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    You're right, looks like a typo in the mark scheme.
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    I actual fact both answers are wrong.

    Remember that you can take multiples outside of integrals:

    \int -\dfrac{3}{2x} dx = -\dfrac{3}{2} \int \dfrac{1}{x} dx

    Then you can intergrate \dfrac{1}{x} and multiply by - \dfrac{3}{2}

             -\dfrac{3}{2}\int \dfrac{1}{x} dx =



     - \dfrac{3}{2} * ln(x) + C =



    -\dfrac{3}{2}ln(x)+C
    • Thread Starter
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    Is this one of those situations where are the above are right but the constant changes to account for the difference?
    Thanks by the way!
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    (Original post by hoodboilu4)
    .
    (Original post by 01Chris02)
    Is this one of those situations where are the above are right but the constant changes to account for the difference?
    Thanks by the way!
    They're not both wrong.
    recall that: -\frac{3}{2}\ln(2x) = -\frac{3}{2}\ln(2) + -\frac{3}{2}\ln(x)
    And also that:
    -\frac{3}{2}\ln(2)
    is just a constant.
    You simply get a different constant (+c) after you integrate.
    • Thread Starter
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    Oh wait sorry I just realised that I typed that in wrong the mark scheme says...
    

-\dfrac{3}{2} \ln (x) +c
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    (Original post by 01Chris02)
    Oh wait sorry I just realised that I typed that in wrong the mark scheme says...
    

-\dfrac{3}{2} \ln (x) +c
    In which case the mark scheme and you are also correct, you would attain full credit
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    (Original post by joostan)
    They're not both wrong.
    recall that: -\frac{3}{2}\ln(2x) = -\frac{3}{2}\ln(2) + -\frac{3}{2}\ln(x)
    And also that:
     = -\frac{3}{2}\ln(2)
    is just a constant.
    You simply get a different constant (+c) after you integrate.
    Thanks for clearing that up!
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    (Original post by 01Chris02)
    Thanks for clearing that up!
    no problem
 
 
 
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