Is $ \int_0 ^1 \frac{dx}{\sqrt{x}}$ an improper integral, even though substituting $u=\sqrt{x}$ turns it into the proper $\frac12\int_0^1du$?

Question

Here is the question,

Is $\;\int_0 ^1 \frac{dx}{\sqrt{x}}\;$ an improper integral?

Of course, that integral is improper because of discontinuity at $x=0$. Nevertheless, after the substitution $u = \sqrt x$ then $du = \dfrac{dx}{2\sqrt x}$.

Therefore, integral turns out to be $$\displaystyle \frac{1}{2}\int_0 ^1 du$$ However, that integral is not improper yet.

$$ \displaystyle \int_0 ^1 \frac{dx}{\sqrt{x}}= \frac{1}{2}\int_0 ^1 du$$

Is there a something wrong or can we generalize that if we have improper integral that converges can turn out to be integral which is not improper?