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Thread: Please check this integral

  1. #1
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    Please check this integral

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

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

    $\displaystyle u+1=^3\sqrt(x)$

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

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

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

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

    $\displaystyle u+1=^3\sqrt(x)$

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

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

    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 $\displaystyle dx$.

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

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

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

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

    $\displaystyle u+1=^3\sqrt(x)$

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

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

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

    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 $\displaystyle \sqrt[3]{x}$. I have enough trouble with people who mistake $\displaystyle \sqrt[3]{3}$ for a square root without Latex reinforcing it! Your "^3\sqrt{x}" is a good substitute.
    Last edited by HallsofIvy; Sep 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 $\displaystyle ^3\sqrt{x}$.
    \sqrt[n]{x} $\displaystyle =\sqrt[n]{x}$

    Last edited by FelixFelicis28; Sep 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
    $\displaystyle \int \frac{^3\sqrt(x)}{^3\sqrt(x)-1}$

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

    $\displaystyle u+1=^3\sqrt(x)$

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

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

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

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

    $\displaystyle \displaystyle \begin{align*} 3\int{ \frac{1}{3\left( \sqrt[3]{x} \right) ^2} \cdot \frac{x}{\sqrt[3]{x} - 1} \, dx} &= 3 \int{ \frac{\left( u + 1 \right) ^3 }{u} \, du} \\ &= 3 \int{ \frac{u^3 + 3u^2 + 3u + 1}{u}\,du} \\ &= 3\int{ u^2 + 3u + 3 + \frac{1}{u}\, du} \\ &= 3 \left( \frac{u^3}{3} + \frac{3u^2}{2} + 3u + \ln{|u|} \right) + C \\ &= u^3 + \frac{9u^2}{2} + 9u + 3\ln{|u|} + C \\ &= \left( \sqrt[3]{x} - 1 \right) ^3 + \frac{9 \left( \sqrt[3]{x} - 1 \right) ^2}{2} + 9 \left( \sqrt[3]{x} - 1 \right) + 3\ln{ \left| \sqrt[3]{x} - 1 \right| } + C \end{align*}$
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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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