I am tasked with determining whether the following improper integral converges: $$\int_0^1\sqrt{\frac{x^2+1}{x}}dx$$Deemed improper as $\lim_{x\to0^+}\sqrt{(x^2+1)/x}=\infty$. After a while of struggling I got Wolfram alpha to tell me that it evaluates to some expression involving hypergeometric functions (whatever they are - but not important!) - i.e: it converges, so it was left to me to demonstrate why.
Seeing as the function doesn't have an elementary antiderivative, it won't be possible to just show $\lim_{x\to0^+}\int_x^1\sqrt{(t^2+1)/t}\;dt$ exists, rather something like comparison to a larger function would be necessary, but I don't really know how to do that to functions in this form. I was also thinking of sort of using a reverse of the integral test for infinite series, by performing some change of variables on the integrand to get something like $\int_1^\infty f(t)dt$ which exists iff $\sum_{n=1}^\infty f(n)$ but I don't know whether (a) this would be valid or (b) this would be any quicker/easier.
Any help regarding this specific integral, or further general methods for dealing with improper integrals, would be much appreciated.
You can use equivalence of functions: near $0$, $x^2+1\sim 1$, so for $x>0 $, $$\sqrt{\frac{x^2+1}{x}}\sim_0\frac1{\sqrt x},$$ and the integral $\;\displaystyle\int_0^1\frac1{\sqrt x}\,\mathrm d x\;$ converges.