Đặt: $\sqrt{1-x}=t \Rightarrow 1-x=t^2 \Rightarrow -dx=2tdt$
Ta có:
$\int\dfrac{1}{x\sqrt{1-x}}dx$
$=-\int\dfrac{2tdt}{(1-t^2)t}$
$=\int\dfrac{2dt}{t^2-1}$
$=\int\left(\dfrac{1}{t-1}-\dfrac{1}{t+1}\right)dt$
$=\ln(t-1)-\ln(t+1)+C$
$=\ln(\sqrt{1-x}-1)-\ln(\sqrt{1-x}+1)+C$