Improper Integral and Comparison Theorem

**adkinsjr** · December 14th 2009, 02:13 PM

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?

**Krizalid** · December 14th 2009, 02:25 PM

for $x\ge1$ it's $\frac{x+1}{\sqrt{x^{4}-x}}>\frac{x}{\sqrt{4x^{4}}}=\frac{1}{2x}.$

**Krizalid** · December 16th 2009, 02:04 AM

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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