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Math Help - Find the indefinite integral

  1. #1
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    Find the indefinite integral

    Hi again

    \int \frac{\sqrt{x}}{x-1}~dx

    it seems none of the u subsitution of  \sqrt{x}, x-1, x works

    some hints would be greatly appreciated =)
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  2. #2
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    Quote Originally Posted by ugkwan View Post
    Hi again

    \int \frac{\sqrt{x}}{x-1}~dx

    it seems none of the u subsitution of  \sqrt{x}, x-1, x works

    some hints would be greatly appreciated =)
    x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)

    \displaystyle\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+1}=\frac{2\sqrt{x}}{x-1}

    Hence you can convert your integral to an integral of the sum of partial fractions
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  3. #3
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    Quote Originally Posted by ugkwan View Post
    Hi again

    \int \frac{\sqrt{x}}{x-1}~dx

    it seems none of the u subsitution of  \sqrt{x}, x-1, x works

    some hints would be greatly appreciated =)
    Try u=\sqrt{x} to get:

    \displaystyle \int \dfrac{2u^2}{u^2-1} \; du

    Now try integration by parts splitting the integrand into 2u and u/(u^2-1)

    CB
    Last edited by CaptainBlack; September 25th 2010 at 01:22 PM. Reason: correct typo, which makes partial fractions an option
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  4. #4
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    Quote Originally Posted by ugkwan View Post
    Hi again

    \int \frac{\sqrt{x}}{x-1}~dx

    it seems none of the u subsitution of  \sqrt{x}, x-1, x works

    some hints would be greatly appreciated =)
    In terms of the substitution, an alternative is

    u=\sqrt{x}

    u^2=x\Rightarrow\frac{d}{dx}u^2=\frac{d}{dx}x\Righ  tarrow\frac{d}{du}u^2\frac{du}{dx}=1\Rightarrow\ 2u\frac{du}{dx}=1

    dx=2udu

    \displaystyle\int{\frac{\sqrt{x}}{x^2-1}dx=\int{\frac{u}{u^2-1}2u}du=\int{\frac{2u^2}{u^2-1}}du

    =\displaystyle\int{\frac{2u^2}{(u+1)(u-1)}}du


    \displaystyle\frac{u}{u-1}+\frac{u}{u+1}=\frac{2u^2}{u^2-1}

    u-1=y\Rightarrow\ du=dy,\;\;\;u=y+1

    w=u+1\Rightarrow\ dw=du,\;\;\;u=w-1

    leaving us with

    \displaystyle\int{\frac{y+1}{y}}dy+\int{\frac{w-1}{w}}dw
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