ozgunozgur
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Can Improper Integrals Help Solve This Inequality?
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skeeter
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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.