Answer
$$(\ln x) \sin^{-1}(\ln x)+\sqrt{1-(\ln x)^2}+c$$
Work Step by Step
Given $$\int \frac{\sin^{-1} (\ln x)}{x}dx$$
Let $s=\ln x\ \Rightarrow ds=\dfrac{1}{x}dx $
then
$$\int \frac{\sin^{-1} (\ln x)}{x}dx=\int \sin^{-1}s ds $$
\begin{align*}
u&=\sin^{-1}s\ \ \ \ \ \ \ \ \ \ \ \ \ \ dv= ds\\
u&=\frac{1}{\sqrt{1-s^2}}ds\ \ \ \ \ \ \ \ \ \ \ \ \ v= s
\end{align*}
\begin{align*}
\int \frac{\sin^{-1} (\ln x)}{x}dx&=\int \sin^{-1}s ds\\
&=s \sin^{-1}s-\int \frac{s}{\sqrt{1-s^2}}ds\\
&=s \sin^{-1}s+\sqrt{1-s^2}+c\\
&=(\ln x) \sin^{-1}(\ln x)+\sqrt{1-(\ln x)^2}+c
\end{align*}
