Answer
$\sqrt{1-(\ln x)^2}+\ln \dfrac{\ln x}{(1+\sqrt{1-(\ln x)^2})}+C$
Work Step by Step
Consider $\ln x=\sin \theta \implies dx/x= \cos \theta d \theta $
Now, the given integral can be written as:
$\int\dfrac{\sqrt{1-\sin^2 \theta}(\cos \theta)}{ \sin \theta} d\theta= \int \csc \theta-\sin \theta d\theta$
or, $=\cos \theta-\ln |\csc d \theta+\cot d \theta|+C$
or, $=\sqrt{1-(\ln x)^2}+\ln \dfrac{\ln x}{(1+\sqrt{1-(\ln x)^2})}+C$
