# Thread: Improper Integral and Comparison Theorem

1. ## Improper Integral and Comparison Theorem

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

By graphing the integrand, it appears that $\displaystyle 0\leq\frac{1}{x}\leq\frac{x+1}{\sqrt{x^4-x}}$ throughout the interval $\displaystyle (1,\infty]$. Since $\displaystyle \int_1^{\infty}\frac{dx}{x}=\lim_{t->\infty}\left(\ln(t)-\ln(1)\right)$ diverges, does this prove that $\displaystyle \int_1^{\infty}\frac{x+1}{\sqrt{x^4-x}}dx$ diverges? Is it that easy?

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

3. i just realized that what i did is wrong.

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

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

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 $\displaystyle x\ge2,$ it still holds and the integral diverges without analyzing the first piece.