Question

${\int }_{\frac{1}{9}}^{\frac{1}{4}}\frac{dx}{(1-x)\sqrt{x}}=\log \frac{3}{2}$

Answer 1

$I={\int }_{1/9}^{1/4}\frac{dx}{(1-x)\sqrt{x}}$

Let $\sqrt{x}=t$ limit , at $\sqrt{1/9}=t=\frac{1}{3}$

$\frac{1}{2\sqrt{x}}dx=dt$ at $x=\frac{1}{4},\sqrt{\frac{1}{4}}=t=\frac{1}{2}$

$\frac{dx}{\sqrt{x}}=2dt$
$\therefore I={\int }_{1/3}^{1/2}\frac{2dt}{(1-t^2)}$

$={\left[\frac{2}{2}\log \left|\frac{1+t}{1-t}\right|\right]}_{1/3}^{1/2}$

$\frac{2}{2}[\log \left|\frac{1+1/2}{1-1/2}\right|-\log \left|\frac{1+1/3}{1-1/3}\right|]$

$=\log \frac{3}{2}$