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Math Help - Imporper Integrals divergence convergence

  1. #1
    Member zangestu888's Avatar
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    Imporper Integrals divergence convergence

    Integral of:
    dx / sqrt (x^4 + 4x^3) from upper limit b=1 and a=0

    Does this integral converge or diverge from the work i did am thinking that it divergers, help someone confirom or show me how it worked!!
    I poseted a simliar thread before same question but with different limits of integration, any help would be appreciated.
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    If it's hard to find an inequality for 0\le x\le1, you may try the following:

    Put x=\frac1u and the integral becomes \int_{1}^{\infty }{\frac{dx}{\sqrt{1+4x}}}\ge \int_{1}^{\infty }{\frac{dx}{\sqrt{5x+4x}}}=\int_{1}^{\infty }{\frac{dx}{3\sqrt{x}}}, which is clearly divergent.
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  3. #3
    Member zangestu888's Avatar
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    I dont quite understand what u did? what happened to my orginal function?
    dx / sqrt (x^4 + 4x^3), can you please tell me the process, and the letting x=1/u am not sure, thanks
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