Welcome to our community

Be a part of something great, join today!

Improper integral question

ozgunozgur

New member
Jun 1, 2020
27

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
638
note ...

$\dfrac{x\sqrt{x}}{x^2-1} > \dfrac{1}{\sqrt{x}}$

and $\displaystyle \int_2^\infty \dfrac{dx}{\sqrt{x}}$ is divergent.