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Thread: Integral

  1. #1
    Junior Member
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    Integral

    Hi,

    How to calculate the following integral $\displaystyle \int_{e^{3}}^{e^{4}}\frac{ln(1+\sqrt{x})}{x+\sqrt{ x}} $ ???
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  2. #2
    Junior Member
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    Start with integration by parts.

    $\displaystyle \int{u*dv}=uv-\int{v*du}$

    $\displaystyle u=ln(1+\sqrt{x})$
    $\displaystyle du=\frac{1}{2\sqrt{x}*(\sqrt{x}+1)}$
    $\displaystyle v=2ln(1+\sqrt{x})$
    $\displaystyle dv=\frac{1}{x+\sqrt{x}}$

    then

    $\displaystyle uv=(ln(1+\sqrt{x}))*(2ln(1+\sqrt{x}))=2ln(1+\sqrt{ x})^2$

    and

    $\displaystyle \int{v*du}=\int{\frac{ln(1+\sqrt{x})}{\sqrt{x}(1+\ sqrt{x}}}$

    Do it again and you'll get

    $\displaystyle \int{\frac{ln(1+\sqrt{x})}{\sqrt{x}(1+\sqrt{x}}}=( integration by parts)=ln(1+\sqrt{x})^2)$

    and so

    $\displaystyle (\int{u*dv}=uv-\int{v*du})=((2ln(1+\sqrt{x})^2)-ln(1+\sqrt{x})^2)))=ln(1+\sqrt{x})^2$

    And of course, now that you have done the integral you can enter the parameters and evaluate it.
    Last edited by shenanigans87; Apr 30th 2010 at 08:17 AM.
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  3. #3
    Super Member
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    $\displaystyle \int \frac{\ln(1+\sqrt{x})}{x+\sqrt{x}}~dx $

    $\displaystyle = \int \frac{1}{\sqrt{x}} \frac{\ln(1+\sqrt{x})}{1+\sqrt{x}}~dx $

    $\displaystyle = 2 \int \frac{\ln(1+\sqrt{x})}{1+\sqrt{x}}~d(\sqrt{x}) $

    $\displaystyle = 2\int \ln(1+\sqrt{x}) ~d(\ln(1+\sqrt{x})) $

    sub. $\displaystyle u = \ln(1+\sqrt{x})$

    The integral

    $\displaystyle = 2 \int udu = u^2 + C = \ln^2(1+\sqrt{x}) + C$
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