- #1
ozgunozgur
- 27
- 0
MHB Can Improper Integrals Help Solve This Inequality?
- Thread starter ozgunozgur
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The inequality $\dfrac{x\sqrt{x}}{x^2-1} > \dfrac{1}{\sqrt{x}}$ is under examination, with a focus on solving it through improper integrals. The integral $\displaystyle \int_2^\infty \dfrac{dx}{\sqrt{x}}$ has been identified as divergent, indicating potential challenges in the analysis. Participants are encouraged to provide assistance in continuing the method outlined. The discussion revolves around the implications of the divergence on the inequality's solution. Overall, the conversation seeks to explore the relationship between improper integrals and the given inequality.
- #2
skeeter
- 1,103
- 1
note ...
$\dfrac{x\sqrt{x}}{x^2-1} > \dfrac{1}{\sqrt{x}}$
and $\displaystyle \int_2^\infty \dfrac{dx}{\sqrt{x}}$ is divergent.
$\dfrac{x\sqrt{x}}{x^2-1} > \dfrac{1}{\sqrt{x}}$
and $\displaystyle \int_2^\infty \dfrac{dx}{\sqrt{x}}$ is divergent.