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Math Help - Another integration problem

  1. #1
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    Another integration problem

    How do I calculate the integral when the denominator is a square root of a function?

    e.g.

    \int \frac{u^3+1}{\sqrt{u^4+4u}} du
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  2. #2
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    This one you can solve using a substitution.

    \int{\frac{u^3 + 1}{\sqrt{u^4 + 4u}}\,du} = \frac{1}{4}\int{(u^4 + 4u)^{-\frac{1}{2}}(4u^3 + 4)\,du}.


    Now let v = u^4 + 4u so that dv = (4u^3 + 4)\,du and the integral becomes

    \frac{1}{4}\int{v^{-\frac{1}{2}}\,dv}

     = \frac{1}{4}\left(\frac{v^{\frac{1}{2}}}{\frac{1}{2  }}\right) + C

     = \frac{1}{4}\left(2\sqrt{v}\right) + C

     = \frac{\sqrt{u^4 + 4u}}{2} + C.
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  3. #3
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    Quote Originally Posted by blackdragon190 View Post
    How do I calculate the integral when the denominator is a square root of a function?
    \int \frac{u^3+1}{\sqrt{u^4+4u}} du
    What is the derivative of \dfrac{\sqrt{u^4+4u}}{2}?
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  4. #4
    MHF Contributor chisigma's Avatar
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    Please note that...

    \displaystyle \frac{u^{3}+1}{\sqrt{u^{4}+4\ u}} = \frac{1}{4}\ \frac{4\ u^{3}+4}{\sqrt{u^{4}+4\ u}} (1)

    ... and that...

    \displaystyle \frac{d}{d u}\ (u^{4}+4\ u) = 4\ u^{3} + 4 (2)

    At this point You are 'almost arrived' ...

    Kind regards

    \chi \sigma
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  5. #5
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    The standard first step is to rationalize the denominator by multiplying the denominator and numerator by the expression sitting in the denominator.
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  6. #6
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    Quote Originally Posted by wonderboy1953 View Post
    The standard first step is to rationalize the denominator by multiplying the denominator and numerator by the expression sitting in the denominator.
    Sorry, but this is not correct.
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  7. #7
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    Thanks! I was just forgetting that a square root in the denominator means that it's to the power of -1/2.
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