Prove that $\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,\text{d}x=-\frac{\pi}{2}\,\ln(2)$.

By letting $x=\sin(t)$, and by using the symmetry $\sin(\pi/2-t)=\cos(t)$, we get $$\begin{align}I&=\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,dx=\int_0^{\pi/2} \ln (\sin t)\, dt =\frac{1}{2}\left(\int_0^{\pi/2} \ln (\sin t)\, dt+\int_0^{\pi/2} \ln (\cos t)\, dt\right)\\&=\frac{1}{2}\left(\int_0^{\pi/2} \ln(\sin(2t))dt - \int_0^{\pi/2} \ln(2)dt\right)=\frac{I}{2}-\dfrac{\pi}4 \ln(2)\end{align}$$ and the result easily follows.

I would start with

$$ J(a) = \int_0^1 \frac{x^a \; dx}{\sqrt{1-x^2}}, \ a > -1$$ The substitution $x = \sqrt{t}$ gives you $$ J(a) = \frac{1}{2} \int_0^1 t^{(a-1)/2} (1-t)^{-1/2}\; dt = \frac{B((a+1)/2,1/2)}{2} = \frac{\Gamma(a+1/2) \Gamma(1/2)}{\Gamma(a+1)} $$ where $B$ is the Beta function. Then your integral is $$J'(0) = - \frac{-\pi \ln(2)}{2}$$

Take $x = \sin t$ then $dx = \cos t \ dt $ so \begin{equation} \frac{\ln (x) }{\sqrt{1 - x^2}} \ dx = \frac{\ln (\sin t) }{\cos t} \cos t \ dt = \ln (\sin t) \ dt \end{equation} SO the integral becomes \begin{equation} A = \int\limits_{0}^{\frac{\pi}{2}} \ln (\sin t) \ dt \end{equation} or just name $x$ instead of $t$ \begin{equation} A = \int\limits_{0}^{\frac{\pi}{2}} \ln (\sin x) \ dx \end{equation} Now use the following change of variable \begin{equation} t = \frac{\pi}{2} - x \end{equation} We get \begin{equation} A = - \int\limits_{\frac{\pi}{2}}^{0} \ln (\sin (\frac{\pi}{2} - t)) \ dt = \int\limits_{0}^{\frac{\pi}{2}} \ln (\cos t) \ dt \end{equation} This means \begin{equation} 2A = \int\limits_{0}^{\frac{\pi}{2}} \ln (\sin x) \ dx + \int\limits_{0}^{\frac{\pi}{2}} \ln (\cos x) \ dx = \int\limits_{0}^{\frac{\pi}{2}} [ \ln (\sin x) \ dx + \ln (\cos x) \ dx ] \end{equation} But $\ln ab = \ln a + \ln b$ so \begin{equation} 2A = \int\limits_{0}^{\frac{\pi}{2}} \ln (\sin x \cos x) \ dx = \int\limits_{0}^{\frac{\pi}{2}} \ln (\frac{1}{2} \sin 2x) \ dx = \int\limits_{0}^{\frac{\pi}{2}} \ln (\frac{1}{2}) + \ln( \sin 2x) \ dx \end{equation} this means that \begin{equation} 2A = \frac{\pi}{2} \ln \frac{1}{2} + B \end{equation} where $B =\int\limits_{0}^{\frac{\pi}{2}} \ln( \sin 2x) \ dx $ \begin{equation} B = \int\limits_{0}^{\frac{\pi}{4}} \ln( \sin 2x) \ dx + \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ln( \sin 2x) \ dx \end{equation} Let $ t = 2x - \frac{\pi}{2}$ so \begin{equation} B = \int\limits_{0}^{\frac{\pi}{4}} \ln( \sin 2t) \ dx + \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ln( \sin 2(t + \frac{\pi}{2})) \ dx = \frac{1}{2} \int\limits_{0}^{\frac{\pi}{2}} \ln(\sin(t)) \ dt + \frac{1}{2} \int\limits_{0}^{\frac{\pi}{2}} \ln(\cos(t)) \ dt \end{equation} which is \begin{equation} B = \frac{1}{2} A + \frac{1}{2} A = A \end{equation} So \begin{equation} 2A = \frac{\pi}{2}\ln \frac{1}{2} + A \end{equation} So \begin{equation} A = \frac{\pi}{2}\ln \frac{1}{2} \end{equation}