A direct proof for $\int_0^x \frac{- x \ln(1-u^2)}{u \sqrt{x^2-u^2}} \, \mathrm{d} u = \arcsin^2(x)$

Let $u=x \sin (\theta)$ \begin{eqnarray*} -\int_0^{\pi/2} \frac{\ln(1-x^2 \sin^2(\theta))}{\sin(\theta)} d \theta \end{eqnarray*} Now expand the logarithms \begin{eqnarray*} \sum_{n=1}^{\infty} \int_0^{\pi/2} \frac{1}{n} x^{2n} \sin^{2n-1}(\theta) d \theta \end{eqnarray*} Now use \begin{eqnarray*} \int_0^{\pi/2} \sin^{2n-1}(\theta) d \theta= \frac{(2n-2)!!}{(2n-1)!!}. \end{eqnarray*} Finally use the result you state in the question \begin{eqnarray*} \frac{1}{2} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} x^{2n}=(\sin^{-1}(x))^2 \end{eqnarray*} and we are done.

I have finally managed to put all the pieces together, so here's a solution that does not use the power series:

Let $u = x v$ to obtain $$ f(x) = \int \limits_0^1 \frac{- \ln(1 - x^2 v^2)}{v \sqrt{1-v^2}} \, \mathrm{d} v \, . $$ Now we can differentiate under the integral sign (justified by the dominated convergence theorem) and use the substitution $v = \sqrt{1 - w^2}\, .$ Then the derivative is given by \begin{align} f'(x) &= 2 x \int \limits_0^1 \frac{v}{(1-x^2 v^2) \sqrt{1-v^2}} \, \mathrm{d} v = 2 x \int \limits_0^1 \frac{\mathrm{d} w }{1-x^2 + x^2 w^2} \\ &= \frac{2}{\sqrt{1-x^2}} \arctan \left(\frac{x}{\sqrt{1-x^2}}\right) = \frac{2 \arcsin (x)}{\sqrt{1-x^2}} \end{align} for $x \in (0,1)$. Since $f(0)=0 \, ,$ integration yields $$ f(x) = f(0) + \int \limits_0^x \frac{2 \arcsin (y)}{\sqrt{1-y^2}} \, \mathrm{d} y = \arcsin^2 (x)$$ for $x \in [0,1]$ as claimed.