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    Please can someone help with 2a and 2b.. I thought I understood this topic but not sure how to do these! Thanks
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    (Original post by jazz_xox_)
    Please can someone help with 2a and 2b.. I thought I understood this topic but not sure how to do these! Thanks
    Find A and B first such that \dfrac{2}{x(x+1)} \equiv \dfrac{A}{x} + \dfrac{B}{x+1} for 2a and \dfrac{4}{(x-1)(x+1)} \equiv \dfrac{A}{x-1} + \dfrac{B}{x+1} for 2b
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    (Original post by RDKGames)
    Find A and B first such that \dfrac{2}{x(x+1)} \equiv \dfrac{A}{x} + \dfrac{B}{x+1} for 2a and \dfrac{4}{(x-1)(x+1)} \equiv \dfrac{A}{x-1} + \dfrac{B}{x+1} for 2b
    Thanks! would you integrate 2/(x+1) by using ln ??
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    (Original post by jazz_xox_)
    Thanks! would you integrate 2/(x+1) by using ln ??
    Yes
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    differentiate ln(x+1) and notice the similarities. as a rule, if you are trying to integrate a fraction where the numerator is a multiple of the derivative of the denominator, you will get a multiple of the natural log of the denominator as the result
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    (Original post by RDKGames)
    Find A and B first such that \dfrac{2}{x(x+1)} \equiv \dfrac{A}{x} + \dfrac{B}{x+1} for 2a and \dfrac{4}{(x-1)(x+1)} \equiv \dfrac{A}{x-1} + \dfrac{B}{x+1} for 2b


    Please could you help with this one too, I'm not sure where I've gone wrong, thanks so much
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    (Original post by jazz_xox_)
    Please could you help with this one too, I'm not sure where I've gone wrong, thanks so much
    You absolutely cannot go from \displaystyle \int \frac{6}{x^2} .dx to \displaystyle \frac{3}{x} \int \frac{2x}{x^2} .dx because x is a variable you're integrating with respect to!
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    (Original post by RDKGames)
    You absolutely cannot go from \displaystyle \int \frac{6}{x^2} .dx to \displaystyle \frac{3}{x} \int \frac{2x}{x^2} .dx because x is a variable you're integrating with respect to!
    Ahh okay, oops! How could I go about integrating this then?

    Thanks so much for all your help, you're such a great help on this forum - and always come here before I ask a teacher

    Edit: would it go to -6x^-1?
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    Got it!

    For the last step,
    when I had 3 - 2ln2 + 2ln5 - 2ln4

    I turned it into this:
    3 -(2ln2 - 2ln5 + 2ln4)
    But why is this necessary to do rather than working with it from the first step? (sorry if it makes no sense)
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    (Original post by jazz_xox_)
    Got it!

    For the last step,
    when I had 3 - 2ln2 + 2ln5 - 2ln4

    I turned it into this:
    3 -(2ln2 - 2ln5 + 2ln4)
    But why is this necessary to do rather than working with it from the first step? (sorry if it makes no sense)
    It’s not necessary. You probably made an error there but not here
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    (Original post by RDKGames)
    It’s not necessary. You probably made an error there but not here
    So how would I get this..

    3 - 2ln2 + 2ln5 - 2ln4

    into

    3 - 2 ln (8/5)

    I did 2ln 5/4 and then 2ln 5/8 (but didn't get the right answer this way?)
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    (Original post by jazz_xox_)
    So how would I get this..

    3 - 2ln2 + 2ln5 - 2ln4

    into

    3 - 2 ln (8/5)

    I did 2ln 5/4 and then 2ln 5/8 (but didn't get the right answer this way?)
    You probably have 3+2\ln \frac{5}{8}. So just swap the sign in front of 2 and flip the fraction inside.
 
 
 
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