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Thread: Need help evaluating indefinate integral.

  1. December 16th 2010, 05:49 AM #1
    saykhong
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    Need help evaluating indefinate integral.

    int [x/(x+1)^(1/2)]
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  2. December 16th 2010, 05:59 AM #2
    TheCoffeeMachine
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    Quote Originally Posted by saykhong View Post
    int [x/(x+1)^(1/2)]
    \int\frac{x}{\sqrt{x+1}}\;{dx}. Letting u = \sqrt{x+1} gives {2}\int\frac{u(u^2-1)}{u}\;{du} = {2}\int\left(u^2-1\right)\;{du} = ...
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  3. December 16th 2010, 06:10 AM #3
    Danny
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    You could also let u = x + 1 then

    \displaystyle \int \dfrac{(u-1)^2}{\sqrt{u}}du expand and integrate.
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  4. December 16th 2010, 03:06 PM #4
    Krizalid
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    \displaystyle\int{\frac{x}{\sqrt{x+1}}\,dx}=\int{\ sqrt{x+1}\,dx}-\int{\frac{dx}{\sqrt{x+1}}}=\frac{2}{3}{{(x+1)}^{\ frac{3}{2}}}-2\sqrt{x+1}+k.
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