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Math Help - Please check this integral

  1. #1
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    Please check this integral

    \int \frac{^3\sqrt(x)}{^3\sqrt(x)-1}

    I let u=^3\sqrt(x)-1

    u+1=^3\sqrt(x)

    \int\frac{u+1}{u}du
    \int (1 + \frac{1}{u})du
    u + ln\abs{u} + c

    ^3\sqrt(x) -1 + ln\abs{^3\sqrt(x)} + c

    Am I close? Wolframalpha comes up with a different method. Thanks.
    How do I get the cube root in Latex?
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  2. #2
    Junior Member FelixFelicis28's Avatar
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    Re: Please check this integral

    Quote Originally Posted by wondering View Post
    \int \frac{^3\sqrt(x)}{^3\sqrt(x)-1}

    I let u=^3\sqrt(x)-1

    u+1=^3\sqrt(x)

    \int\frac{u+1}{u}du
    \int (1 + \frac{1}{u})du
    u + ln\abs{u} + c

    ^3\sqrt(x) -1 + ln\abs{^3\sqrt(x)} + c

    Am I close? Wolframalpha comes up with a different method. Thanks.
    How do I get the cube root in Latex?
    You have forgetten to make a substitution for dx.

    u = x^{1/3} - 1 \implies \frac{du}{dx} = 1/3 x^{-2/3} \implies dx = du 3(u+1)^2

    Btw, the LaTeX for the cube root is \sqrt[3]{x}
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  3. #3
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    Re: Please check this integral

    Quote Originally Posted by wondering View Post
    \int \frac{^3\sqrt(x)}{^3\sqrt(x)-1}

    I let u=^3\sqrt(x)-1

    u+1=^3\sqrt(x)

    \int\frac{u+1}{u}du
    Okay, if u= ^3\sqrt{x}- 1 then u+ 1= ^3\sqrt{x} and so \frac{^3\sqrt{x}}{^3\sqrt{x}- 1}= \frac{u+ 1}{u} but what happened to the "dx"? if ^3\sqrt{x}= u+ 1 then x= (u+ 1)^3= u^3+ 3u^2+ 3u+ 1 so [tex]dx= (3u^2+ 6u+ 3)du
    \int \frac{^3\sqrt{x}}{^3\sqrt{x}- 1}dx= \int \frac{u+1}{u}(3u^2+ 6u+ 3)du= \int (3u^2+ 6u+ 3+ 3u+ 6+ \frac{3}{u})du= \int (3u^2+ 9u+ 9+ \frac{3}{u})du

    \int (1 + \frac{1}{u})du
    u + ln\abs{u} + c

    ^3\sqrt(x) -1 + ln\abs{^3\sqrt(x)} + c

    Am I close? Wolframalpha comes up with a different method. Thanks.
    How do I get the cube root in Latex?
    The best way to get a cube root (though I dislike it my self!) is "\sqrt[3]{x}" gives \sqrt[3]{x}. I have enough trouble with people who mistake \sqrt[3]{3} for a square root without Latex reinforcing it! Your "^3\sqrt{x}" is a good substitute.
    Last edited by HallsofIvy; September 21st 2013 at 05:32 PM.
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  4. #4
    Junior Member FelixFelicis28's Avatar
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    Re: Please check this integral

    Quote Originally Posted by HallsofIvy View Post
    What you did is the best way to get a cube root (though I dislike it my self!) "^3\sqrt{x}" gives ^3\sqrt{x}.
    \sqrt[n]{x} =\sqrt[n]{x}

    Last edited by FelixFelicis28; September 21st 2013 at 05:31 PM.
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  5. #5
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    Re: Please check this integral

    Darn! I was hoping to edit my post so I wouldn't look so foolish but I wasn't fast enough!
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  6. #6
    Junior Member FelixFelicis28's Avatar
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    Re: Please check this integral

    Just too quick.
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  7. #7
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    Re: Please check this integral

    Quote Originally Posted by wondering View Post
    \int \frac{^3\sqrt(x)}{^3\sqrt(x)-1}

    I let u=^3\sqrt(x)-1

    u+1=^3\sqrt(x)

    \int\frac{u+1}{u}du
    \int (1 + \frac{1}{u})du
    u + ln\abs{u} + c

    ^3\sqrt(x) -1 + ln\abs{^3\sqrt(x)} + c

    Am I close? Wolframalpha comes up with a different method. Thanks.
    How do I get the cube root in Latex?
    \displaystyle \begin{align*} \int{\frac{\sqrt[3]{x}}{\sqrt[3]{x} - 1}\,dx} &= \int{\frac{\left( \sqrt[3]{x} \right) ^2 \sqrt[3]{x} }{\left( \sqrt[3]{x} \right) ^2 \left( \sqrt[3]{x} - 1 \right) } \, dx} \\ &= 3\int{ \frac{1}{3\left( \sqrt[3]{x} \right) ^2} \cdot \frac{x}{\sqrt[3]{x} - 1} \, dx} \end{align*}

    Now make the substitution \displaystyle \begin{align*} u = \sqrt[3]{x} - 1 \implies du = \frac{1}{3\left( \sqrt[3]{x} \right) ^2 } \, dx \end{align*} and note that this substitution also means \displaystyle \begin{align*} x = \left( u + 1 \right) ^3 \end{align*}, and the integral becomes

    \displaystyle \begin{align*} 3\int{ \frac{1}{3\left( \sqrt[3]{x} \right) ^2} \cdot \frac{x}{\sqrt[3]{x} - 1} \, dx} &= 3 \int{ \frac{\left( u + 1 \right) ^3 }{u} \, du} \\ &= 3 \int{ \frac{u^3 + 3u^2 + 3u + 1}{u}\,du} \\ &= 3\int{ u^2 + 3u + 3 + \frac{1}{u}\, du} \\ &= 3 \left( \frac{u^3}{3} + \frac{3u^2}{2} + 3u + \ln{|u|} \right) + C \\ &= u^3 + \frac{9u^2}{2} + 9u + 3\ln{|u|} + C \\ &= \left( \sqrt[3]{x} - 1 \right) ^3 + \frac{9 \left( \sqrt[3]{x} - 1 \right) ^2}{2} + 9 \left( \sqrt[3]{x} - 1 \right) + 3\ln{ \left| \sqrt[3]{x} - 1 \right| } + C \end{align*}
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  8. #8
    Forum Admin topsquark's Avatar
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    Re: Please check this integral

    Y'all could simply use x^{1/3}...

    -Dan
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  9. #9
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    Re: Please check this integral

    Thanks for all of the quick replies, it makes alot more sense now.
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