Integrate the following: $\int \frac {\log x}{\sqrt {1-x^2}}\,dx$

I do not think that we could find a closed form expression except using very special and complex function.

If $$I=\int \frac {\log (x)}{\sqrt {1-x^2}}dx$$ just as you did using integration by parts $$u=\log(x)\implies du=\frac{dx}x$$ $$dv=\frac {dx}{\sqrt {1-x^2}}\implies v=\sin^{-1}(x)$$ then $$I=\log (x) \sin ^{-1}(x)-\int \frac{\sin ^{-1}(x)}{x}\,dx$$ This last integral could be computed using Taylor series $$\sin^{-1}(x)=\sum^{\infty}_{n=0} \frac{(2n)!}{4^n (n!)^2 (2n+1)} x^{2n+1}\quad\text{ for }|x| \le 1$$ which makes, integrting each term of the expansion, $$\int \frac{\sin ^{-1}(x)}{x}\,dx=\sum^{\infty}_{n=0} \frac{(2n)!}{4^n (n!)^2 (2n+1)^2} x^{2n+1}$$

Just for your curiosity, this last summation corresponds to $$\int \frac{\sin ^{-1}(x)}{x}\,dx=x \,\, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};x^2\right )$$ where appears the generalized hypergeometric function.

What is amazing (at least to me !) is that this monster is quite "close" to $\frac{\pi \log (2)}{2}x $.

Edit

Another possible way is to start with the expansion $$\frac {1}{\sqrt {1-x^2}}=\sum_{n=0}^ \infty \frac{\binom{2 n}{n} }{4^n}x^{2n}$$ That is to say that $$I=\int \frac {\log (x)}{\sqrt {1-x^2}}=\sum_{n=0}^ \infty \frac{\binom{2 n}{n} }{4^n\,(2n+1)^2}\left((2n+1)\log(x)-1 \right)x^{2n+1}$$ $$I=x (\log (x)-1) \, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};x^2\right )+\frac{1}{9} x^3 \log (x) \, _3F_2\left(\frac{3}{2},\frac{3}{2},\frac{3}{2};\frac{5}{2},\frac{5}{2};x^2\right )$$