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Math Help - find or evaluate the integral

  1. #1
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    find or evaluate the integral

    Hi

    \displystyle \int \frac{1}{1 + \sqrt{x}}~dx

    I have tried multiplying the numerator with +\sqrt{x} and -\sqrt{x} to factor out the denominator and got stuck
    I have tried substituted 1 + \sqrt{x} and got stuck
    I have tried substituting \sqrt{x} with u and replaced the numerator with  u - \sqrt{x} and came up with nothing...

    Insights please.
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by ugkwan View Post
    Hi

    \displystyle \int \frac{1}{1 + \sqrt{x}}~dx

    I have tried multiplying the numerator with +\sqrt{x} and  -\sqrt{x} to factor out the denominator and got stuck
    I have tried substituted 1 + \sqrt{x} and got stuck
    I have tried substituting  \sqrt{x} with u and replaced the numerator with  u - \sqrt{x} and came up with nothing...

    Insights please.


    use  u = \sqrt {x} so  du = \frac {1}{2\sqrt{x}}

    so now your integral will be

     \displaystyle 2\int \frac {u}{u+1} du = \displaystyle 2 \int (1 - \frac {1}{1+u} ) du



    can you continue from here ?
    Last edited by yeKciM; September 25th 2010 at 10:38 AM.
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  3. #3
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    so what is the antiderivative of \frac {u}{u+1}?
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  4. #4
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    nevermind my previous question. I figured out that by adding 1 and minusing 1 to the numerator, i can split the function, and then proceed with the antiderivative process.
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