Does $\int_{0}^{1}\frac{\sqrt{x}}{\ln{2x}}dx$ diverge or converge? Not really sure how to compare this one. It's clearly undefined in it's lower bound and in $x=\frac{1}{2}$. If $x\in[0,1]$, it follows that
$$\int_{0}^{1}\frac{\sqrt{x}}{\ln{2x}}dx\leq\int_{0}^{1}\frac{1}{\ln{2x}}dx=\int_{0}^{1/2}\frac{1}{\ln{2x}}dx+\int_{1/2}^{1}\frac{1}{\ln{2x}}dx.$$
But this did not make anything easier, since both of the last integrals can't be expressed in elementary functions.
NOTE: I'm not allowed to solve this with Taylor's L'Hopitals or anything similar.
 A: When in doubt, remove the logarithm through a suitable substitution, it is rarely a bad idea.
Here, by letting $x=\frac{1}{2}e^t$, we have
$$ I = \int_{0}^{1}\frac{\sqrt{x}}{\log(2x)}\,dx =\frac{1}{2\sqrt{2}}\int_{-\infty}^{\log 2}\frac{e^{3t/2}}{t}\,dt $$
and $\frac{1}{2\sqrt{2}}\int_{-\infty}^{-\log 2}\frac{e^{3t/2}}{t}\,dt $ is a harmless, convergent integral (by the Cauchy-Schwarz inequality, if you like). The troubling part comes with
$$ \int_{-\log 2}^{\log 2}\frac{e^{3t/2}}{t}\,dt $$
which, strictly speaking, does not exist due to the non-integrable singularity at the origin. On the other hand such integral is convergent in a regularized sense:
$$\text{PV}\int_{-\log 2}^{\log 2}\frac{e^{3t/2}}{t}\,dt=\int_{0}^{\log 2}\frac{2\sinh\frac{3t}{2}}{t}\,dt\approx\int_{0}^{\log 2}\frac{3t}{t}\,dt=3\log 2 $$
since $\sinh$ is almost linear in a right neighbourhood of the origin (and for a more precise inequality, one may exploit its convexity). So the answer to your question "is XXX convergent?" strongly depends on what you mean by convergent, which should be clarified, together with the allowed or forbidden tools. 
A: The integral diverges due to the singularity at $x=1/2$ where $\log(2x)=\log(1+2(x-1/2))=2(x-1/2)+O(x-1/2)^2$.  
Hence, the integrand behaves like $\frac1{x-1/2}$ locally around $x=1/2$ and the integral fails to exist.
We can define, however, a Cauchy Principal Value integral as
$$PV\int_0^1 \frac{\sqrt x}{\log(2x)}\,dx=\lim_{\epsilon\to 0^+}\left(\int_0^{1/2-\epsilon}\frac{\sqrt x}{\log(2x)}\,dx+\int_{1/2+\epsilon}^1 \frac{\sqrt x}{\log(2x)}\,dx\right)$$
which does converge.
