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    I just can't get the integration part of the question right:


    a)express 2x+1/(2x+3)^2 in the form A/2x+3 + B/(2x+3)^2


    I got 1/2x+3 - 2/(2x+3)^2


    b) hence show that (the integral between 2 and -1) of 2x+1/(2x+3)^2 dx = 1/2ln7 - 6/7

    I know that you integrate the partial fractions but i think i've gone wrong somewhere after that
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    (Original post by mica-lwe)
    I just can't get the integration part of the question right:


    a)express 2x+1/(2x+3)^2 in the form A/2x+3 + B/(2x+3)^2


    I got 1/2x+3 - 2/(2x+3)^2


    b) hence show that (the integral between 2 and -1) of 2x+1/(2x+3)^2 dx = 1/2ln7 - 6/7

    I know that you integrate the partial fractions but i think i've gone wrong somewhere after that
    post workings please
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    (Original post by TeeEm)
    post workings please
    ^^^there they are
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    (Original post by mica-lwe)
    ^^^there they are
    2nd fraction does not integrate to a log
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    (Original post by TeeEm)
    2nd fraction does not integrate to a log
    how do you integrate it then?
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    (Original post by mica-lwe)
    how do you integrate it then?
    write it to the minus 2 and integrate by inspection(recognition)


    or use substitution
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    (Original post by TeeEm)
    write it to the minus 2 and integrate by inspection(recognition)


    or use substitution
    to integrate a bracket do you raise the power by one, then divide the bracket by the new power x the differential of the bracket?
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    (Original post by mica-lwe)
    to integrate a bracket do you raise the power by one, then divide the bracket by the new power x the differential of the bracket?
    yes

    (so long as the differential of the bracket is a constant, which is the case here)
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    (Original post by TeeEm)
    yes

    (so long as the differential of the bracket is a constant, which is the case here)
    I almost have the right answer, I have 2ln7 -6/7 but I need 1/2ln7 -6/7, don't know why it's 1/2 instead of 2?
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    (Original post by mica-lwe)
    I almost have the right answer, I have 2ln7 -6/7 but I need 1/2ln7 -6/7, don't know why it's 1/2 instead of 2?
    after your partial fractions you have put a 2 in front of the integral

    when you integrated the log, you have not divided by 2


    hopefully when you sort them it will work
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    (Original post by TeeEm)
    after your partial fractions you have put a 2 in front of the integral

    when you integrated the log, you have not divided by 2


    hopefully when you sort them it will work
    Thankyou!
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    (Original post by mica-lwe)
    Thankyou!
    you are welcome
 
 
 
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