Answer
\[\frac{1}{2}\ln\left|(\ln x)^2+\sqrt{(\ln x)^4+1}\right|+C\]
Work Step by Step
Let \[I=\int\frac{\ln x}{x\sqrt{1+(\ln x)^2}}dx\;\;\;\ldots(1)\]
Substitute $t=(\ln x)^2$ ____(2)
$\;\;\;\;\;\;dt=\frac{2(\ln x)}{x}dx$
\[I=\frac{1}{2}\int\frac{dt}{\sqrt{1+t^2}}\]
\[\left[\; \int\frac{dx}{\sqrt{x^2+a^2}}=\ln\left|x+\sqrt{x^2+a^2}\right|\;\right]\]
\[I=\frac{1}{2}\ln|t+\sqrt{t^2+1}|+C\]
Where $C$ is constant of integration
From (2)
\[I=\frac{1}{2}\ln\left|(\ln x)^2+\sqrt{(\ln x)^4+1}\right|+C\]
Hence,
$I=\large\frac{1}{2}$ $\ln\left|(\ln x)^2+\sqrt{(\ln x)^4+1}\right|+C$.
