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Math Help - 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:

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

    t:=ln(1+\sqrt{x}) \Rightarrow dt=\frac{1}{2\sqrt{x}(1+\sqrt{x})} <br />
\Rightarrow dx=2\sqrt{x}(1+\sqrt{x})dt
    \Downarrow
    e^t=1+\sqrt{x} \Rightarrow x=(e^t-1)^2

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

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

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

    = \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:

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

    t:=ln(1+\sqrt{x}) \Rightarrow dt=\frac{1}{2\sqrt{x}(1+\sqrt{x})} <br />
\Rightarrow dx=2\sqrt{x}(1+\sqrt{x})dt
    \Downarrow
    e^t=1+\sqrt{x} \Rightarrow x=(e^t-1)^2

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

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

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

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

    Now what ?

    Thank you very much
    Click on Show steps: integrate Sqrt&#x5b;x&#x5d;Log&#x5b;1 &#x2b; Sqrt&#x5b;x&#x5d;&#x5d; - Wolfram|Alpha
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  3. #3
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    Let u = \sqrt{x} + 1 so your integral becomes

     <br />
2 \int (u-1)^2 \ln u \, du<br />
.

    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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