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    can anyone help me? im useless

    integrate x+1 / (1-x^2) ^1/2 dx

    in simple english
    integrate x+1 divided by 1-xsquared to the half with respect to x!


    anyone?
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    (Original post by kriztinae)
    can anyone help me? im useless

    integrate x+1 / (1-x^2) ^1/2 dx

    in simple english
    integrate x+1 divided by 1-xsquared to the half with respect to x!


    anyone?
    INT (x+1) / (1-x^2) ^1/2 dx

    = INT x/(1-x^2)^1/2 + 1/(1-x^2)^1/2 dx

    For the first one sub u = 1 - x^2 if you want or recognise derivative second one is just arcsinx
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    (Original post by It'sPhil...)
    INT (x+1) / (1-x^2) ^1/2 dx

    = INT x/(1-x^2)^1/2 + 1/(1-x^2)^1/2 dx

    For the first one sub u = 1 - x^2 if you want or recognise derivative second one is just arcsinx
    ok thanx
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    (Original post by It'sPhil...)
    INT (x+1) / (1-x^2) ^1/2 dx

    = INT x/(1-x^2)^1/2 + 1/(1-x^2)^1/2 dx

    For the first one sub u = 1 - x^2 if you want or recognise derivative second one is just arcsinx
    hey can anyone help me on this same question? im stuck integrating the first part
    ok to put it simpler!
    how do you integrate:
    x/(1-x^2)1/2 dx

    anyone please? any help much apreciated thanx
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    >how do you integrate:
    >x/(1-x^2)1/2 dx

    as mentioned b4 use u = 1-x^2

    so du/dx = 2x

    du/2 = x dx

    now x dx is what u have for your numerator.....so

    Int 1/(u^1/2) du/2

    1/2 * Int u^(-1/2) du

    now its easy

    mrm.
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    (Original post by Mrm.)
    >how do you integrate:
    >x/(1-x^2)1/2 dx

    as mentioned b4 use u = 1-x^2

    so du/dx = 2x

    du/2 = x dx

    now x dx is what u have for your numerator.....so

    Int 1/(u^1/2) du/2

    1/2 * Int u^(-1/2) du

    now its easy

    mrm.
    i dont see it
    dont worry about it ill find another way! thanx anyway
 
 
 
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