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    Can someone show me how to integrate (x^-2 -1)^-1/2
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    Do as you normally do: add 1 to the power and divide by the new power. Then divide by the differntial of the bracket, you can check your answer by differentiating.

    You should get

    (-2/2x)(x^-2 -1)^1/2
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    Multiply top and bottom by x:

    \displaystyle \frac{1}{\sqrt{\frac{1}{x^2}-1}} = \frac{x}{\sqrt{x^2}\sqrt{\frac{1  }{x^2}-1}} = \frac{x}{\sqrt{1-x^2}}

    Now notice that the derivative of (1-x^2) is 2x.
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    also, how would you integrate (x^2 - 2x)^-1/2
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    (Original post by buckett)
    Do as you normally do: add 1 to the power and divide by the new power. Then divide by the differntial of the bracket, you can check your answer by differentiating.

    You should get

    (-2/2x)(x^-2 -1)^1/2
    I don't know how much integration you have done but this is the wrong method. Most of the time, the rules of differentiation cannot be reversed when integrating.

    Try differentiating your answer and see what you get.
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    (Original post by notnek)
    Multiply top and bottom by x:

    \displaystyle \frac{1}{\sqrt{\frac{1}{x^2}-1}} = \frac{x}{\sqrt{x^2}\sqrt{\frac{1  }{x^2}-1}} = \frac{x}{\sqrt{1-x^2}}

    Now notice that the derivative of (1-x^2) is 2x.
    ye i can get that far, but your forgetting that the bottom is rooted?
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    (Original post by Robofish)
    ye i can get that far, but your forgetting that the bottom is rooted?
    I'm not forgetting that. Try the substitution u=1-x^2 and you'll see that it works.

    In general,

    \displaystyle \int \frac{k f'(x)}{f(x)^n} = k\frac{f(x)^{1-n}}{1-n}

    where f is a function and f' is it's derivative.
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    (Original post by notnek)
    I'm not forgetting that. Try the substitution u=1-x^2 and you'll see that it works.

    In general,

    \displaystyle \int \frac{k f'(x)}{f(x)^n} = k\frac{f(x)^{1-n}}{1-n}

    where f is a function and f' is it's derivative.
    oh ye it does work, im such an idiot loll is there a similar substitution for the other integral i posted?
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    (Original post by Robofish)
    oh ye it does work, im such an idiot loll is there a similar substitution for the other integral i posted?
    I'm not really sure about the other integral. I'm a bit rusty so I recommend posting the integral in a new thread so other people can have a go.
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    Actually, I do now how to do it.

    Complete the square in the denominator and then substitute u=(x-1)
 
 
 
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Updated: February 7, 2010
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