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Math Help - Help Integrate: sqrt(2+x^(1/2))

  1. #1
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    Help Integrate: sqrt(2+x^(1/2))

    I cannot figure out how to integrate this expression. A root of a root? Trig substitution does not work. My ti 89 says answer is
    (4/15)(sqrt(x)+2)(3sqrt(x)-4). Any help is appreciated.
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  2. #2
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    Hello, speedengine!

    \int \sqrt{2 + \sqrt{x}}\:dx

    Answer: . \frac{4}{15}(\sqrt{x}+2)^{\frac{3}{2}}(3\sqrt{x} - 4) + C

    With "nested" radicals, it often works if we let u equal something inside.

    Let: 2 + \sqrt{x}\:=\:u\quad\Rightarrow\quad \sqrt{x}\:=\:u-2\quad\Rightarrow\quad x u - 2)^2\quad\Rightarrow\quad dx \:=\:2(u-2)\,du" alt=" \:=\u - 2)^2\quad\Rightarrow\quad dx \:=\:2(u-2)\,du" />

    Substitute: . \int u^{\frac{1}{2}}\cdot2(u - 2)\,du\;=\;2\int\left(u^{\frac{3}{2}} - 2u^{\frac{1}{2}}\right)\,du \;=\;2\left(\frac{2}{5}u^{\frac{5}{2}} - \frac{4}{3}u^{\frac{3}{2}}\right) + C

    Factor out \frac{2}{15}:\;\;\frac{4}{15}\left(3u^{\frac{5}{2}  } - 10u^{\frac{3}{2}}\right) + C

    Factor out u^{\frac{3}{2}}:\;\;\frac{4}{15}u^{\frac{3}{2}}(3u - 10) + C


    Since u\;=\;\sqrt{x} + 2, we have: . \frac{4}{15}(\sqrt{x} + 2)^{\frac{3}{2}}\left[3(\sqrt{x} + 2) - 10\right] + C

    . . . = \;\frac{4}{15}(\sqrt{x} + 2)^{\frac{3}{2}}(3\sqrt{x} + 6 - 10) + C\;=  \;\boxed{\frac{4}{15}(\sqrt{x} + 2)^{\frac{3}{2}}(3\sqrt{x} - 4) + C}

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  3. #3
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    Hey Soroban,
    Awesome! Defining x in terms of u is where I was missing. Thanks a bunch!
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