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Thread: Integrals - ln, sqrts

  1. #1
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    Integrals - ln, sqrts

    My English-Math language isn't too good, so I'll try to use math to explain:

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

    $\displaystyle t:=ln(1+\sqrt{x}) \Rightarrow dt=\frac{1}{2\sqrt{x}(1+\sqrt{x})}
    \Rightarrow dx=2\sqrt{x}(1+\sqrt{x})dt$
    $\displaystyle \Downarrow$
    $\displaystyle e^t=1+\sqrt{x} \Rightarrow x=(e^t-1)^2$

    $\displaystyle \int \sqrt{x} \ln{(1+\sqrt{x})} dx$ = $\displaystyle \int \sqrt{x} t*2\sqrt{x}(1+\sqrt{x})dt$

    = $\displaystyle \int 2t(x+x^\frac{3}{2}) dt$

    =$\displaystyle \int 2t((e^t-1)^2+(e^t-1)^3) dt $

    =$\displaystyle \int 2t*e^t(e^t-1)^2)) dt $

    Now what ?

    Thank you very much
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  2. #2
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    Quote Originally Posted by adam63 View Post
    My English-Math language isn't too good, so I'll try to use math to explain:

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

    $\displaystyle t:=ln(1+\sqrt{x}) \Rightarrow dt=\frac{1}{2\sqrt{x}(1+\sqrt{x})}
    \Rightarrow dx=2\sqrt{x}(1+\sqrt{x})dt$
    $\displaystyle \Downarrow$
    $\displaystyle e^t=1+\sqrt{x} \Rightarrow x=(e^t-1)^2$

    $\displaystyle \int \sqrt{x} \ln{(1+\sqrt{x})} dx$ = $\displaystyle \int \sqrt{x} t*2\sqrt{x}(1+\sqrt{x})dt$

    = $\displaystyle \int 2t(x+x^\frac{3}{2}) dt$

    =$\displaystyle \int 2t((e^t-1)^2+(e^t-1)^3) dt $

    =$\displaystyle \int 2t*e^t(e^t-1)^2)) dt $

    Now what ?

    Thank you very much
    Click on Show steps: integrate Sqrt[x]Log[1 + Sqrt[x]] - Wolfram|Alpha
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  3. #3
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    Let $\displaystyle u = \sqrt{x} + 1$ so your integral becomes

    $\displaystyle
    2 \int (u-1)^2 \ln u \, du
    $.

    Expand and integrate by parts.
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  4. #4
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    Thanks for the help I got it!
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