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Math Help - Integration Proglem

  1. #1
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    Cool Integration Proglem

    I'm trying to integrate the function 1/(1 + x^1/2) dx. Any help would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by kaiser0792 View Post
    I'm trying to integrate the function 1/(1 + x^1/2) dx. Any help would be greatly appreciated.
    Let u=\sqrt{x} to get:

    2\int \frac{u}{u+1} du.
    Can you complete it?
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kaiser0792 View Post
    I'm trying to integrate the function 1/(1 + x^1/2) dx. Any help would be greatly appreciated.
    Let u = \sqrt x

    \Rightarrow u^2 = x

    \Rightarrow 2u~du = dx

    So your integral becomes

    2 \int \frac u{1 + u}~du

    = 2 \int \left( \frac {u + 1 - 1}{u + 1} \right)~du

    = 2 \int \left( 1 - \frac 1{u + 1} \right)~du

    I suppose the rest will be easy for you
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  4. #4
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    Thanks General

    That did it! I missed squaring u and adding and subtracting 1.

    I really appreciate the help, it was driving me crazy!!
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  5. #5
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    let u = x^1/2, u^2 = x and then 2u du = dx.

    Rewriting: 2*integral (u/u+1) du = 2*integral [u+1-1]/(u+1)du
    = 2[integral (u+1)/(u+1)du - integral 1/(u+1)du]
    = 2[integral du - integral 1/(u+1)du] Let v = u+1, dv=du
    = 2[u - integral dv/v = 2u - ln|v|] + C
    = 2[x^1/2 - ln|u+1|] + C = 2[x^1/2 - ln|(x^1/2 + 1)|] + C
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  6. #6
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    Quote Originally Posted by kaiser0792 View Post
    let u = x^1/2, u^2 = x and then 2u du = dx.

    Rewriting: 2*integral (u/u+1) du = 2*integral [u+1-1]/(u+1)du
    = 2[integral (u+1)/(u+1)du - integral 1/(u+1)du]
    = 2[integral du - integral 1/(u+1)du] Let v = u+1, dv=du
    = 2[u - integral dv/v = 2u - ln|v|] + C
    = 2[x^1/2 - ln|u+1|] + C = 2[x^1/2 - ln|(x^1/2 + 1)|] + C
    Its correct.
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  7. #7
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    Thanks for your help and for your follow up.
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