I need help evaluating this with contour integration $$ \int_{0}^{1}{\ln\left(\,x\,\right)\over \,\sqrt{\vphantom{\large A}\,1 - x^{2}\,}}\,{\rm d}x $$ I am not sure as to how to work with the branch cuts of both $\ln\left(\,x\,\right)$ and $\sqrt{1 - x^{2}}$
Second part is to evaluate $$ \int_{0}^1 \frac{\sqrt{\,\vphantom{\large a}\ln\left(\,x\,\right)}} {\sqrt{\vphantom{\large A}\,1 - x^{2}\,}} \mathrm dx$$
For your second integral, we can use Ramanujan's method which I mentioned here. That is, let $x = e^{-t}$. Then we have \begin{align} \int_0^1\frac{\log^{1/2}(x)}{\sqrt{1-x^2}}dx &= \int_0^{\infty}\frac{ie^{-t}\sqrt{t}}{\sqrt{1-e^{-2t}}}dt\\ &=i\int_0^{\infty}\sqrt{t}\sum_{n=0}^{\infty}\frac{(1/2)_ne^{-(2n+1)t}}{n!}dt\tag{1}\\ &=i\sum_{n=0}^{\infty}\frac{(1/2)_n}{n!}\int_0^{\infty}t^{1/2}e^{-(2n+1)t}dt\tag{2}\\ &=\frac{i\sqrt{\pi}}{2}\sum_{n=0}^{\infty}\frac{(1/2)_n}{n!(2n+1)^{3/2}} \end{align} $(1)$ occurs because $\sum_{n=0}^{\infty}\frac{(1/2)_ne^{-(2n+1)t}}{n!}=\frac{e^{-t}}{\sqrt{1-e^{-2t}}}$ and $(1/2)_n$ is the rising Pochhammer; that is, $(1/2)_0 = 1$ and $(1/2)_n = 1/2(1/2+1)\cdots(1/2+n-1)$. For $(2)$, we have that $$ \int_0^{\infty}t^{1/2}e^{-(2n+1)t}dt = \frac{\Gamma(3/2)}{(2n+1)^{3/2}}=\frac{\sqrt{\pi}}{2(2n+1)^{3/2}} $$
As one of the comments suggested, we do want to obtain $\ln(\sin(\theta))$. Then we will solve the problem using complex analysis.
Let $x = \sin(\theta)$ and $dx = \cos(\theta)d\theta$. Therefore, $0\mapsto 0$ and $1\mapsto\pi/2$. Thus, the integral is $$ \int_0^{\pi/2}\ln(\sin(\theta))d\theta = \frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta $$ Consider $1 - e^{2iz} = -2ie^{iz}\sin(z)$. We can write $1 - e^{2iz}$ as $$ 1 - e^{-2y}(\cos(2x) + i\sin(2x)) < 0\text{ when } x=\pi n, \ y\leq 0 $$ Now let's consider the contour from $0$ to $\pi$ to $\pi + iA$ to $iA$ where we take a quarter of a circle around $0$ and $\pi$ with radius $\epsilon$. From the periodicity of the function, the vertical line segments cancel each other since they have opposite signs. Additionally, as $A\to\infty$, the top integral of the top line goes to zero and as $\epsilon\to 0$, the integral around $0$ and $\pi$ go to zero. \begin{align} \ln(-2ie^{ix}\sin(z)) &= \ln(-2i) + \ln(e^{ix}) + \ln(\sin(\theta))\\ &= \ln|-2i| + i\arg(-2i) + ix + \ln(\sin(\theta))\\ &= \ln(2) - i\frac{\pi}{2} + \ln(\sin(\theta)) + i\frac{\pi}{2} \end{align} where $\ln(2i) = \ln(2) + i\arg(-2i)$ and we take the principle argument to be $-\frac{\pi}{2}$ and the imaginary part of $ix$ is between $0$ and $\pi$. \begin{alignat}{2} \int_0^1\frac{\ln(x)}{\sqrt{1-x^2}}dx &= \int_0^{\pi/2}\ln(\sin(\theta))d\theta\\ &=\frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta\\ &= \frac{\ln(2)}{2}\int_0^{\pi}d\theta - \frac{i\pi}{4}\int_0^{\pi}d\theta + \frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta + \frac{i\pi}{4}\int_0^{\pi}d\theta &&{}= 0\\ &= \frac{\pi\ln(2)}{2} - \frac{i\pi^2}{4} + \frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta + \frac{i\pi^2}{4} &&{}=0\\ \frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta &= -\frac{\pi\ln(2)}{2}\\ \int_0^{\pi/2}\ln(\sin(\theta))d\theta &= -\frac{\pi\ln(2)}{2} \end{alignat}
