Answer
$\displaystyle \frac{\ln x}{2}\sqrt{4+(\ln x)^{2}}+2\ln[\ln x+\sqrt{4+(\ln x)^{2}}]+C$
Work Step by Step
$\displaystyle \int\frac{\sqrt{4+(\ln x)^{2}}}{x}dx=\quad$substitute $\left[\begin{array}{ll}
u=\ln x & \\
du=\dfrac{dx}{x} &
\end{array}\right]$
$=\displaystyle \int\sqrt{2^{2}+u^{2}}du=$
Table of integrals:
$\color{blue}{ 21. \quad \displaystyle \int\sqrt{a^{2}+u^{2}}du=\frac{u}{2}\sqrt{a^{2}+u^{2}}+\frac{a^{2}}{2}\ln(u+\sqrt{a^{2}+u^{2}})+C }$
$=\displaystyle \frac{u}{2}\sqrt{4+u^{2}}+\frac{4}{2}\ln(u+\sqrt{4+u^{2}})+C$
... bring back x...
$=\displaystyle \frac{\ln x}{2}\sqrt{4+(\ln x)^{2}}+2\ln[\ln x+\sqrt{4+(\ln x)^{2}}]+C$
