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Math Help - integral (sqr{3}x + sqr(x)) / (sqr{6}x + 1)

  1. #1
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    integral (sqr{3}x + sqr(x)) / (sqr{6}x + 1)

    \int \frac{\sqrt[3]{x} + \sqrt{x}}{\sqrt[6]{x} + 1} dx

    and also

    \int \frac{x+8}{(x-2)(x^2+1)} dx
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  2. #2
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by lainmaster View Post
    \int \frac{\sqrt[3]{x} + \sqrt{x}}{\sqrt[6]{x} + 1} dx
    \frac{\sqrt[3]x+\sqrt x}{\sqrt[6]x+1}

    =\frac{\left(\sqrt[3]x+\sqrt x\right)\left(\sqrt[6]x-1\right)}{\left(\sqrt[6]x+1\right)\left(\sqrt[6]x-1\right)}

    =\frac{x^{1/2}-x^{1/3}+x^{2/3}-x^{1/2}}{x^{1/3}-1}

    =\frac{x^{2/3}-x^{1/3}}{x^{1/3}-1}

    =\frac{x^{1/3}\left(x^{1/3}-1\right)}{x^{1/3}-1}

    =x^{1/3}

    Now integrate.

    \int \frac{x+8}{(x-2)(x^2+1)} dx
    Use partial fraction decomposition.

    Edit: And make sure you are familiar with the integral \int\!\frac{dx}{1+x^2}.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Reckoner View Post
    \frac{\sqrt[3]x+\sqrt x}{\sqrt[6]x+1}

    =\frac{\left(\sqrt[3]x+\sqrt x\right)\left(\sqrt[6]x-1\right)}{\left(\sqrt[6]x+1\right)\left(\sqrt[6]x-1\right)}

    =\frac{x^{1/2}-x^{1/3}+x^{2/3}-x^{1/2}}{x^{1/3}-1}

    =\frac{x^{2/3}-x^{1/3}}{x^{1/3}-1}

    =\frac{x^{1/3}\left(x^{1/3}-1\right)}{x^{1/3}-1}

    =x^{1/3}

    Now integrate.
    very nice!

    using your final answer as motivation, we could have just factored out x^{1/3} from the numerator in the beginning. then everything else cancels in the next step. but seeing that that was a good move in the first place is probably hard, i guess.
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  4. #4
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    Thanks, Reckoner

    EDIT

    It seems so simple now... but how did you come up with that so easily? Is there a method, or you just saw it crystal clear from so much practice?
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