integral (from 0 to infinity) of dx/rootx(x+1)
When x is large, the integrand is approximately 1/x, so the integral will diverge.
If that explanation isn't rigorous enought for you, you can integrate $\displaystyle \int\frac{dx}{\sqrt{x(x+1)}} = \int\frac{2dx}{\sqrt{(2x+1)^2-1}}$ by making the substitution $\displaystyle 2x+1=\sec\theta$.