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Math Help - Find the Definite Integral

  1. #1
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    Find the Definite Integral

    Hi

    (intetral sign) 1/ (1+ 2^(1/2)) dx

    I subsituted u= 1 + 2^(1/2) but got no where after du = x^(-1/2) dx

    Algebraic help or hints please.
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  2. #2
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    \displaystyle{\frac{1}{1 + 2^{\frac{1}{2}}}} is a constant. What does that tell you about its antiderivative?
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  3. #3
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    Maybe have a closer look at your quesiton as,

    \displaystyle \int \frac{1}{1+2^{\frac{1}{2}}}~dx =\frac{x}{1+2^{\frac{1}{2}}}+C
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  4. #4
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    I am so sorry that I incorrectly posted the problem. I should of taken the time to double check instead of wasting your time.

    The question is ∫ 1/ (1 + √(2x) )

    Again I substituted 1 + √(2x) and got stuck with du = 1/√(x)
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  5. #5
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    \frac{1}{1 + \sqrt{2x}} = \frac{\sqrt{x}}{\sqrt{x}(1 + \sqrt{2x})}

     = \frac{\sqrt{x}}{1 + \sqrt{2}\sqrt{x}}\left(\frac{1}{\sqrt{x}}\right)

     = \frac{2\sqrt{x}}{1 + \sqrt{2}\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\right).


    Therefore

    \int{\frac{1}{1 + \sqrt{2x}}\,dx} = \int{\frac{2\sqrt{x}}{1 + \sqrt{2}\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\right)\  ,dx}.

    Now let u = \sqrt{x} so that \frac{du}{dx} = \frac{1}{2\sqrt{x}} so that

    \int{\frac{2\sqrt{x}}{1 + \sqrt{2}\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\right)\  ,dx} = \int{\frac{2u}{1 + \sqrt{2}u}\,\frac{du}{dx}\,dx}

     = \int{\frac{2u}{1 + \sqrt{2}u}\,du}

     = 2\int{\frac{u}{1 + \sqrt{2}u}\,du}.


    Now long divide to get

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

     = 2\left[\frac{u}{\sqrt{2}} - \frac{1}{2}\ln{(1 + \sqrt{2}u)}\right] + C

     = \sqrt{2}u - \ln{(1 + \sqrt{2}u)} + C

     = \sqrt{2}\sqrt{x} - \ln{(1 + \sqrt{2}\sqrt{x})} + C

     = \sqrt{2x} - \ln{(1 + \sqrt{2x})} + C.
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  6. #6
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    Shorter way: let \displaystyle u = 1+\sqrt{2x} = 1+\sqrt{2}\sqrt{x}, then \frac{du}{dx} = \frac{\sqrt{2}}{2\sqrt{x}} \Rightarrow {dx} = \frac{2\sqrt{x}}{\sqrt{2}}\;{du}  .
    We have then \displaystyle \frac{2}{\sqrt{2}}\int \frac{\sqrt{x}}{u}\;{du}. But from u = 1+\sqrt{2x} = 1+\sqrt{2}\sqrt{x}, we get \sqrt{x} = \frac{u-1}{\sqrt{2}}.
    So we have \displaystyle \frac{2}{\sqrt{2}\sqrt{2}}\int\frac{u-1}{u}\;{du} = \int\left(1-\frac{1}{u}\right)\;{du} = u-\ln{u}+k.

    As \displaystyle u = 1+\sqrt{2x}, this is \displaystyle 1+\sqrt{2x}-\ln(1+\sqrt{2x})+k.
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