Originally Posted by

**larson** Use the comparison test to determine wehter the improper integral converges or diverges:

$\displaystyle \int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)}$

So, since this is true... $\displaystyle \int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)}$ < $\displaystyle \int_{0}^{\infty} \frac{1}{\sqrt{x}}$ I can just use the larger integral to find out whether or not it converges or diverges because it is much simpler.

$\displaystyle \int_{0}^{\infty} x^\frac{-1}{2}$ which equals:

$\displaystyle \lim_{t \to \infty} \int_{0}^{t} x^\frac{-1}{2}$ equals

$\displaystyle \lim_{t \to \infty} \frac {x^\frac{1}{2}}{2}$ from 0 -> $\displaystyle \infty$

So this just turns out to be $\displaystyle \infty - \frac {0^\frac{1}{2}}{2}$ so this is equal to...

$\displaystyle \infty - 0$ which = $\displaystyle \infty$

So because this integral DIVERGES TO $\displaystyle \infty$, this tells us nothing? So what do I do now?