I worked out $\int _1^{\infty }\frac{\sqrt{4x+5}}{x^2}\:dx$ to have an undefined limit, which tells me it diverges. I'm still trying to fully understand improper integrals so somebody correct me if this integral doesn't diverge.
I wanted to ask, if an improper integral equals some value L, does that imply it converges?
Edit: How do you use the direct comparison test to confirm whether my improper integral actually diverges?
The answer to your first question is "yes." That's what convergence means.
With experience, you'd look at your integrand and sort of blur out the constants and think "that's about $\sqrt{x}/x^2 = 1/x^{3/2}$ and the integral of that converges, so my original integral converges."
Then to prove it with the direct comparison test, you'd try to find a multiple of $1/x^{3/2}$ that is larger than your integrand. Something like:
$$\frac{\sqrt{4x+5}}{x^2} < \frac{\sqrt{10x}}{x^2} = \frac{\sqrt{10}}{x^{3/2}}$$
since for $x\geq 1$ you have $10x \geq 4x+5.$
Because $$\int_1^\infty \frac{\sqrt{10}}{x^{3/2}} \; dx$$
converges, so does your original integral.
If an improper indefinite integral equals to some value or rather expression $L$ then it does not necessarily mean that after putting the limits you'll get a value.
In your case, first evaluate the indefinite integral of the same. It is equal to $$-\dfrac{2\left(\ln\left(\sqrt{4x+5}+\sqrt{5}\right)-\ln\left(\left|\sqrt{4x+5}-\sqrt{5}\right|\right)\right)}{\sqrt{5}}-\dfrac{\sqrt{4x+5}}{x}$$ Now when you'll put the limits $1,\infty$ you'll get the following result $$3-\dfrac{2\ln\left(-\frac{3\sqrt{5}-7}{2}\right)}{\sqrt{5}}$$
But try putting the limits $0,\infty$ then you'll see that you can't put the limits as the integral will behave divergent in this case.
