Does $\int_{1}^{2}\frac{dx}{\sqrt{(2-x)\ln(x)}}$ converge? I have improper integral $$\displaystyle \int_{1}^{2}\frac{dx}{\sqrt{(2-x)\ln(x)}}$$ and I need to find out if it converges/diverges. Now I don't if my thinking is correct: 
$\displaystyle \int \frac{dx}{\sqrt{\ln(x)}}$ goes to $0$ slower than $\displaystyle \int \frac{dx}{\sqrt{x}}$, thus from integral comparison: $$\displaystyle \int_{1}^{2}\frac{dx}{\sqrt{(2-x)\ln(x)}} < \int_{1}^{2} \frac{dx}{\sqrt{x}}$$ it diverges. Is this correct? Thanks.

Bounding above the integral by a convergent one does not prove the integral diverges. 
Actually it converges. The integrand is non-negative and


*

*at $2$, we have $\frac 1{(2-x)\log (x)}\sim \frac 1{(2-x)\log 2}$, and the integral $\int_{3/2}^{2}\frac{\mathrm dx}{\sqrt{2-x}}$ is convergent;

*at $1$, using $\log(1+t)\overset{t\to 0}{\sim}t$, we have $\frac 1{(2-x)\log (x)}\sim \frac 1{\log(x-1+1)}\sim \frac 1{x-1}$.

