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Integrating Brackets

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    How do you integrate brackets.

    For example, (x-1)^-2

    any help appreciated.
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    substitution? Ie, let u=x-1
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    u = x-1

    du/dx = 1

    dx = du

    integral u^-2 du

    = -u^-1
    = -(x-1)^-1
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    Use the chain rule. Add one to the power, then divide the whole bracket by (the new power times by the dericitive of whats inside the bracket)....Thus giving you (x-1)^-1/-1
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    (Original post by theflyingpig)
    u = x-1

    du/dx = 1

    dx = du

    integral u^-2 du

    = -u^-1
    = -(x-1)^-1
    You're right.
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    (Original post by buj)
    How do you integrate brackets.

    For example, (x-1)^-2

    any help appreciated.
    If you don't want to use substitution, add one to the power to give  (x-1)^{-1} . If you then differentiate back again, you get  - (x-1)^{-2} , which is -1 times too big. We therefore need to divide what we differentiated by -1, giving the answer as  -(x-1)^{-1} .
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    (Original post by 0129Hippy)
    You're wrong.
    Nope you need to revise. -2 + 1 does not equal -3
  8. Offline

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    (Original post by 0129Hippy)
    You're wrong.
    No he's right.
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    Ooooopsies...I apologise will all my heart.
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    First two people were wrong.

    You get 1/(1-x)

    By doing it the way Daniel Friedman explained.
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    (Original post by buj)
    How do you integrate brackets.

    For example, (x-1)^-2

    any help appreciated.
    you've left it quite late :eek:

    past papers are your friends
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    (Original post by uthred50)
    Actually, you're both wrong.

    You get 1/(1-x)
    No you're wrong lol

    -(x-1)^-1 = 1/(1-x)
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    (Original post by n1r4v)
    No you're wrong lol

    -(x-1)^-1 = 1/(1-x)
    i was refering to the other dude, but like 3 people posted before i put that post lol

    The answer is definately 1/(1-x), whcih can also be phrased as (1-x)^-1
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    you can not integrate brackets....first of all you need to expand the braket... and then integrate.
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    (Original post by medione)
    you can not integrate brackets....first of all you need to expand the braket... and then integrate.
    You are wrong, unless you want to be forever integrating an infinite expansion....
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    (Original post by medione)
    you can not integrate brackets....first of all you need to expand the braket... and then integrate.
    Yes you can, substitution or chain rule.
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    If your revising for Edexcel Core 4 on Thursday you really should be able to do this by substitution or by observation by now.
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    we're all wrong. the answer is \displaystyle \pi...
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    (Original post by louiscbrooks)
    If your revising for Edexcel Core 4 on Thursday you really should be able to do this by substitution or by observation by now.
    Do you have any more logic puzzles :p:
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    (Original post by n1r4v)
    Do you have any more logic puzzles :p:
    lol no

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