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Math Help - Improper Integral and Comparison Theorem

  1. #1
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    Improper Integral and Comparison Theorem

    I'm trying to use the comparison theorem to show that \int_1^{\infty}\frac{x+1}{\sqrt{x^4-x}}dx is divergent.

    By graphing the integrand, it appears that 0\leq\frac{1}{x}\leq\frac{x+1}{\sqrt{x^4-x}} throughout the interval (1,\infty]. Since \int_1^{\infty}\frac{dx}{x}=\lim_{t->\infty}\left(\ln(t)-\ln(1)\right) diverges, does this prove that \int_1^{\infty}\frac{x+1}{\sqrt{x^4-x}}dx diverges? Is it that easy?
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  2. #2
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    for x\ge1 it's \frac{x+1}{\sqrt{x^{4}-x}}>\frac{x}{\sqrt{4x^{4}}}=\frac{1}{2x}.
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  3. #3
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    i just realized that what i did is wrong.

    note that my inequality doesn't hold for x=1, that's a another case which need to be treated appart.

    we split the interval into two: (1,2] and [2,\infty).

    in order to have convergence, we require the convergence for both integral, but to get a divergence, we just need that one of them be divergent, so, if you use my estimation now for x\ge2, it still holds and the integral diverges without analyzing the first piece.
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