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Math Help - tough integral?.

  1. #1
    Eater of Worlds
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    tough integral?.

    Here's an integral to try. I couldn't even get Maple to give me a closed form for it. Kriz will probably conquer it like Grant through Richmond.

    \int\frac{\sqrt{1+x^{2}}}{\sqrt{10-x^{3}}}dx
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  2. #2
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    Quote Originally Posted by galactus View Post
    Here's an integral to try. I couldn't even get Maple to give me a closed form for it. Kriz will probably conquer it like Grant through Richmond.

    \int\frac{\sqrt{1+x^{2}}}{\sqrt{10-x^{3}}}dx
    My guess would be to start thusly:

    \int\frac{\sqrt{1+x^2}}{\sqrt{10-x^3}}dx

    \int\frac{\sqrt{1+x^2}}{\sqrt{10-x^3}}\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}<br />
dx

    \int\frac{1+x^2}{\sqrt{(10-x^3)(1+x^2)}}dx

    \int\frac{1}{\sqrt{(10-x^3)(1+x^2)}}dx+\int\frac{x^2}{\sqrt{(10-x^3)(1+x^2)}}dx

    How to finish I cannot tell immediately, or even whether the above is an appropriate beginning.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    Here's an integral to try. I couldn't even get Maple to give me a closed form for it. Kriz will probably conquer it like Grant through Richmond.

    \int\frac{\sqrt{1+x^{2}}}{\sqrt{10-x^{3}}}dx
    I almost got it by letting

    u=\sqrt{1+x^2}

    which results in something nasty that you put a trig sub in for and it looks easy...it is but then WHAM! not anymore...try it for yourself =(
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  4. #4
    Math Engineering Student
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    Mmm, no, I think there's no antiderivative for this one.

    But the user NonCommAlg is better than me. He may give an opinion too.
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