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Thread: partial integration problem

  1. #1
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    partial integration problem

    Hello!

    i have difficulties sovling the following equation:

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

    the actual integration part isnt the problem:

    $\displaystyle ln(x)*2\sqrt{x} \int x^\frac{-2}{3} dx$

    simplifying the integral: $\displaystyle \int x^\frac{-3}{2}$

    and after integrating im here:


    $\displaystyle ln(x)*2\sqrt{x} * (-\frac{2}{\sqrt(x)})+C$

    the solution is supposed to be $\displaystyle 2*ln(x)*\sqrt(x)-4\sqrt(x)+C$

    i dont know how to get this result
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by coobe View Post
    Hello!

    i have difficulties sovling the following equation:

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

    the actual integration part isnt the problem:

    $\displaystyle ln(x)*2\sqrt{x} \int x^\frac{-2}{3} dx$

    simplifying the integral: $\displaystyle \int x^\frac{-3}{2}$

    and after integrating im here:


    $\displaystyle ln(x)*2\sqrt{x} * (-\frac{2}{\sqrt(x)})+C$

    the solution is supposed to be $\displaystyle 2*ln(x)*\sqrt(x)-4\sqrt(x)+C$

    i dont know how to get this result

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

    so let

    $\displaystyle u=\ln(x) \implies du =\frac{1}{x}dx$
    $\displaystyle dv=\frac{1}{\sqrt{x}}dx=x^{-1/2}dx \implies v =2\sqrt{x}dx$

    $\displaystyle \int \frac{ln(x)} {\sqrt{x}} dx=2\ln(x) \sqrt{x}-2\int x^{-1/2}dx$

    $\displaystyle \int \frac{ln(x)} {\sqrt{x}} dx=2\ln(x) \sqrt{x}-4\sqrt{x}+c$
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by coobe View Post
    Hello!

    i have difficulties sovling the following equation:

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

    the actual integration part isnt the problem:

    $\displaystyle ln(x)*2\sqrt{x} \int x^\frac{-2}{3} dx$

    simplifying the integral: $\displaystyle \int x^\frac{-3}{2}$

    and after integrating im here:


    $\displaystyle ln(x)*2\sqrt{x} * (-\frac{2}{\sqrt(x)})+C$

    the solution is supposed to be $\displaystyle 2*ln(x)*\sqrt(x)-4\sqrt(x)+C$

    i dont know how to get this result

    let

    $\displaystyle lnx=u...x=e^u...dx=e^udu...x^{\frac{1}{2}}=e^{\fra c{u}{2}}$

    $\displaystyle \int \frac{u(e^u)}{e^{\frac{u}{2}}} du$$\displaystyle =\int u(e^{u-\frac{u}{2}}) du $

    $\displaystyle \int ue^{\frac{u}{2}} du = $

    by parts$\displaystyle dy=e^{\frac{u}{2}} ...y=2e^{\frac{u}{2}}...and...t=u...dt=du$

    $\displaystyle 2ue^{\frac{u}{2}}-2\int e^{\frac{u}{2}} du $

    $\displaystyle 2ue^{\frac{u}{2}} -2(2)e^{\frac{u}{2}} + c $

    sub $\displaystyle u=ln(x).......e^{\frac{u}{2}}=\sqrt{x} $

    someone is faster than me lol
    Last edited by Amer; Jun 10th 2009 at 07:41 AM.
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  4. #4
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    Hello, coobe!

    I'm not sure what you did . . .


    $\displaystyle \int \frac{\ln x} {\sqrt{x}}\,dx$
    We have: .$\displaystyle I \;=\;\int\ln x\left(x^{-\frac{1}{2}}\,dx\right)$

    Integrate by parts: .$\displaystyle \begin{array}{ccccccc}
    u &=&\ln x & & dv &=& x^{-\frac{1}{2}}dx \\ du &=& \frac{dx}{x} & & v &=& 2x^{\frac{1}{2}} \end{array}$

    Then: .$\displaystyle I \;\;=\;\;2\!\cdot\!x^{\frac{1}{2}}\!\cdot\!\ln x - \int\left(2x^{\frac{1}{3}}\right)\left(\frac{dx}{x }\right)$

    . . . . . $\displaystyle =\;\;2\!\cdot\!x^{\frac{1}{2}}\!\cdot\!\ln x - 2\int x^{-\frac{1}{2}}\,dx $

    . . . . . $\displaystyle = \;\;2\cdot\ln x\cdot\!x^{\frac{1}{2}} - 4x^{\frac{1}{2}} + C$


    . . . . . $\displaystyle = \;\;2\cdot\ln x\!\cdot\!\sqrt{x} - 4\sqrt{x} + C$

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  5. #5
    Junior Member
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    d'oh.... i had a wrong formula on my Formula Sheet... took me so much time to figure that out

    thanks everybody, i simply had a wrong formula for partial integrating...

    good thing i noticed that before the exam
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