Very much interesting

Let us first substitute $\ln(x)=t$

So we have the following integral:-

$\displaystyle \int_{\ln(2)}^{\ln(3)}\frac{1}{\sqrt{\left(t-\ln(2)\right)\left(\ln(3)-t\right)}}\,dt$

Now I would like to prove a result of the Beta Function :-

Let us consider the integral

$\displaystyle \int_{a}^{b}(x-a)^{m-1}(b-x)^{n-1}dx$

Substitute $\displaystyle y=\frac{x-a}{b-a}$ .

We get:-

$\displaystyle (b-a)^{m+n-1}\int_{0}^{1}y^{m-1}(1-y)^{n-1}\,dy = (b-a)^{m+n-1}\Beta\left(m,n\right)$ where $\Beta(m,n)$ denotes the beta function

So in the integral given in the question we have $a=\ln(2)$ , $b=\ln(3)$ and $m=n=\frac{1}{2}$ .

So we have the answer as :- $\displaystyle (\ln(3)-\ln(2))^{\frac{1}{2}+\frac{1}{2}-1}\Beta(\frac{1}{2},\frac{1}{2}) = \Beta(\frac{1}{2},\frac{1}{2}) = \frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{2})}{\Gamma(1)}$

Here $\Gamma(.)$ denotes the Gamma Function and we know that $\Gamma(\frac{1}{2})=\sqrt{\pi}$

So our answer is $\displaystyle \Gamma(\frac{1}{2})\Gamma(\frac{1}{2})= \pi \approx 3.1415$

Some u-substitution.....enjoy !