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Math Help - Need help integrating this

  1. #1
    Member zzizi's Avatar
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    Need help integrating this

    Hi can someone help me integrate this:

    \int \frac{1}{4}x^{2}\frac{1}{\sqrt{x^{3}+1}}dx

    I must have forgotten, but do I use u substitution?
    Last edited by zzizi; December 2nd 2012 at 01:21 PM.
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  2. #2
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    Re: Need help integrating this

    Yes, start with  u = x^3, du = 3x^2 dx
    This becomes
     \frac{1}{12} \int \frac{1}{\sqrt{u} +1} du

    To integrate this im not really sure off the top of my head how to do it formally

     = \frac{1}{12}( 2\sqrt{u} -2log(\sqrt{u}+1))

    err had an arithmetic error.
    Last edited by Scopur; December 2nd 2012 at 01:18 PM.
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  3. #3
    Member zzizi's Avatar
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    Re: Need help integrating this

    Thank you very much scopur!

    I'm really sorry but I just realised that the 1 was meant to be under the square-root symbol. I have made he changes, hopefully that doesn't that affect anything too much?
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  4. #4
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    Re: Need help integrating this

    With practice, this sort of problem is solved by differentiation rather than integration.
    Look at the x^{2} on the top line and the \sqrt{x^{3}+1} on the bottom line and realise that they have come from differentiating \sqrt{x^{3}+1}.
    So, differentiate \sqrt{x^{3}+1}=(x^{3}+1)^{1/2} function of a function fashion to get (1/2)(x^{3}+1)^{-1/2}.3x^{2} and then multiply by a suitable constant to get the required result.
    With just a little practice, you just write down the result.

    If you prefer the integration by substitution, u^{2}=x^{3}+1 works best.
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  5. #5
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    Re: Need help integrating this

    Are you sure you'd let \displaystyle \begin{align*} u^2 = x^3 + 1 \end{align*}? I think \displaystyle \begin{align*} u = x^3 + 1 \implies du = 3x^2 \, dx \end{align*} works easier...

    \displaystyle \begin{align*} \int{\frac{1}{4}\, x^2\, \frac{1}{\sqrt{x^3 + 1}}\,dx} &= \frac{1}{12} \int{ 3x^2\, \frac{1}{\sqrt{x^3 + 1}}\, dx } \\ &= \frac{1}{2} \int{\frac{1}{\sqrt{u}}\,du} \\ &= \frac{1}{2} \int{u^{-\frac{1}{2}}\,du}  \end{align*}

    Go from here...
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  6. #6
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    Re: Need help integrating this

    If u^{2}=x^{3}+1, then 2u\, du = 3x^{2}dx so

    \int \frac{1}{4}\frac{x^{2}}{\sqrt{x^{3}+1}}\,dx=\frac{  1}{12}\int \frac{1}{\sqrt{x^{3}+1}}\,3x^{2}\,dx=\frac{1}{12} \int \frac{1}{u}\,2u\,du = \frac{1}{12} \int 2 \, du, etc..
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