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Math Help - Rigorous Integration

  1. #1
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    Rigorous Integration

    \int(\sqrt((x+1)/(x-2))<br />
    Last edited by Cbarker1; June 5th 2012 at 12:12 PM.
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  2. #2
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    Re: Rigorous Integration

    Quote Originally Posted by Cbarker1 View Post
    \int(\sqrt((x+1)/(x-2))
    integral sqrt((x+1)/(x-2)) dx - Wolfram|Alpha

    click on "show steps" ...
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    Re: Rigorous Integration

    I am confusion how that could be legal in the math world.
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    Re: Rigorous Integration

    Quote Originally Posted by Cbarker1 View Post
    I am confusion how that could be legal in the math world.
    I am confusion how that could be legal in the math world.

    Please tell us what that sentence could possibly mean.
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    Re: Rigorous Integration

    I have difficulty written sentences. But i know what i am talking about.
    Sorry.
    I rewrite that sentence to a questions.

    Why the steps in Wolfram Alpha could be correct? If yes, how?
    Last edited by Cbarker1; June 5th 2012 at 12:48 PM.
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  6. #6
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    Re: Rigorous Integration

    Wolfram Alpha suggests using the substitution u= \frac{x+1}{x- 1} precisely what is inside the square root. With that, the integrand becomes just \sqrt{u}= u^{1/2}. Now, what about dx?

    Well, if u= \frac{x+1}{x-1} then u(x- 1)= ux- u= x+ 1 so ux- x= u+ 1, x(u- 1)= u+1, and x= \frac{u+1}{u- 1}. From that, dx= \frac{(u-1)- (u+1)}{(u-1)^2}du= -\frac{2}{(u-1)^2}du.

    Then \int\sqrt{\frac{x+1}{x-1}}dx= \int u^{1/2}\left(-\frac{2}{(u-1)^2}\right)du
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