Answer
$\ln x\cdot \arcsin(\ln x) +\sqrt{1-(\ln x)^{2}} +C$
Work Step by Step
Substitute $\left[\begin{array}{ll}
t=\ln x, & x=e^{t}\\
dt=\frac{dx}{x} &
\end{array}\right]$
$I=\displaystyle \int\frac{\arcsin(\ln x)}{x}dx=\int\arcsin tdt$
By parts, we take $\displaystyle \left[\begin{array}{ll}
u=\arcsin t & dv=dt\\
& \\
du=\frac{1}{\sqrt{1-t^{2}}} & v=t
\end{array}\right],\quad\int udv=uv-\int vdu$
$I=t\displaystyle \arcsin t-\int\frac{t}{\sqrt{1-t^{2}}}dt$
$I_{1}=\displaystyle \int\frac{tdt}{\sqrt{1-t^{2}}}$= substitute $\left[\begin{array}{l}
w=1-t^{2}\\
dw=-2tdt
\end{array}\right]$
= $-\displaystyle \frac{1}{2}\int w^{-1/2}dt=-\frac{1}{2}(\frac{x^{1/2}}{1/2})$
$=-w^{1/2}+C$
$=-\sqrt{1-t^{2}}$
$I=t\arcsin t+\sqrt{1-t^{2}}+C$
... bring back x...
$I=\ln x\cdot \arcsin(\ln x) +\sqrt{1-(\ln x)^{2}} +C$
