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Thread: integrating a function

  1. #1
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    integrating a function

    integrate xlnx/(square root x^2-1)
    how i started:
    by using integration by parts i got (lnx)(2/3(x^2-1)^(3/2) - integral 2/3(x^2-1)^(3/2)/(x)
    where i let u= lnx v=2/3(x^2-1)^3/2
    du= (1/x)dx dv= x(x^2-1)^-(1/2)dx
    I dont how to take the integral of that function.
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Check the formula you used and than the integration you performed Is it correct
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  3. #3
    Like a stone-audioslave ADARSH's Avatar
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    I think this might help you better
    integration by parts means
    $\displaystyle
    \int{u \times v dx} = u\int{vdx}-\int{(u'\int{vdx}) dx}

    $

    So for your function lets take $\displaystyle u= ln(x)$ and $\displaystyle
    v=\frac{x}{\sqrt{x^2-1}}

    $

    Hence the integral becomes

    $\displaystyle

    \int{\frac{x~ln(x)dx}{\sqrt{x^2-1} } } = ln(x) \int{\frac{xdx}{\sqrt{x^2-1}} }- \int{((\frac{d}{dx} ln(x))\int{\frac{xdx}{\sqrt{x^2-1}}})dx}
    $


    First of all
    $\displaystyle
    \int{\frac{x~dx}{\sqrt{x^2-1}}}

    $

    Put $\displaystyle x^2-1 = t$

    Hence $\displaystyle xdx =dt/2$

    Hence the integral becomes
    $\displaystyle \int{\frac{2~dt}{\sqrt{t}}}$

    $\displaystyle
    =\frac{2\sqrt{t}}{2} + c
    $

    And

    $\displaystyle \frac{d}{dx} ln(x) =\frac{1}{x} $

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